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GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

Real Numbers Polynomials Pair of Linear Equations in Two Variables Quadratic Equations Arithmetic Progressions Triangles Coordinate Geometry Introduction to Trigonometry Some Applications of Trigonometry Circles Constructions Areas related to Circles Surface Areas and Volumes Statistics Probability
If ∆ABC ~ ∆PQR, perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then find the length of AC. (2012)

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∆ABC ~ ∆DEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm, find the perimeter of ∆DEF. (2012, 2017D)

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જવાબ : ∆ABC – ∆DEF …[Given


If ∆ABC ~ ∆RPQ, AB = 3 cm, BC = 5 cm, AC = 6 cm, RP = 6 cm and PQ = 10, then find QR. (2014)

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In ∆DEW, AB || EW. If AD = 4 cm, DE = 12 cm and DW = 24 cm, then find the value of DB. (2015)

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જવાબ : Let BD = x cm
then BW = (24 – x) cm, AE = 12 – 4 = 8 cm
In ∆DEW, AB || EW


In ∆ABC, DE || BC, find the value of x. (2015)

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જવાબ : In ∆ABC, DE || BC …[Given

x(x + 5) = (x + 3)(x + 1)
x2 + 5x = x2 + 3x + x + 3
x2 + 5x – x2 – 3x – x = 3
∴ x = 3 cm


In the given figure, if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find DE. (2014)

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જવાબ : In ∆ADE and ∆ABC,
∠DAE = ∠BAC …Common
∠ADE – ∠ABC … [Corresponding angles
∆ADE – ∆ΑΒC …[AA corollary


In the given figure, XY || QR, PQXQ=73 and PR = 6.3 cm, find YR. (2017OD)

 

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જવાબ : Let YR = x
PQXQ=PRYR … [Thales’ theorem


If PQR is an equilateral triangle and PX ⊥ QR, find the value of PX2. (2013)

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જવાબ : Altitude of an equilateral ∆,


The sides AB and AC and the perimeter P, of ∆ABC are respectively three times the corresponding sides DE and DF and the perimeter P, of ∆DEF. Are the two triangles similar? If yes, find ar(△ABC)ar(△DEF) (2012)

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જવાબ : AB = 3DE and AC = 3DF

…[∵ The ratio of the areas of two similar ∆s is equal to the ratio of the squares of their corresponding sides


X and Y are points on the sides AB and AC respectively of a triangle ABC such that AXAB=14, AY = 2 cm and YC = 6 cm. Find whether XY || BC or not. (2015)

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જવાબ :


In the figure, EF || AC, BC = 10 cm, AB = 13 cm and EC = 2 cm, find AF. (2014)

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જવાબ : BE = BC – EC = 10 – 2 = 8 cm
Let AF = x cm, then BF = (13 – x) cm
In ∆ABC, EF || AC … [Given


Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?
 

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જવાબ : Since the perimeters and two sides are proportional
∴ The third side is proportional to the corresponding third side.
i.e., The two triangles will be similar by SSS criterion.


A and B are respectively the points on the sides PQ and PR of a ∆PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm, and PB = 4 cm. Is AB || QR? Give reason.3

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જવાબ :


If ∆ABC ~ ∆QRP, ar(ΔABC)ar(ΔPQR) = 94, AB = 18 cm and BC = 15 cm, then find the length of PR.

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જવાબ :


If it is given that ∆ABC ~ ∆PQR with BCQR = 13, then find ar(ΔPQR)ar(ΔABC)

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જવાબ :


∆DEF ~ ∆ABC, if DE : AB = 2 : 3 and ar(∆DEF) is equal to 44 square units. Find the area (∆ABC).

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Is the triangle with sides 12 cm, 16 cm and 18 cm a right triangle? Give reason.

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જવાબ : Here, 122+ 162 = 144 + 256 = 400 ≠ 182
∴ The given triangle is not a right triangle.


In triangles PQR and TSM, ∠P = 55°, ∠Q = 25°, ∠M = 100°, and ∠S = 25°. Is ∆QPR ~ ∆TSM? Why?
 

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જવાબ : Şince, ∠R = 180° – (∠P + ∠Q)
= 180° – (55° + 25°) = 100° = ∠M
∠Q = ∠S = 25° (Given)
∆QPR ~ ∆STM
i.e., . ∆QPR is not similar to ∆TSM.


If ABC and DEF are similar triangles such that ∠A = 47° and ∠E = 63°, then the measures of ∠C = 70°. Is it true? Give reason.

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જવાબ : Since ∆ABC ~ ∆DEF
∴ ∠A = ∠D = 47°
∠B = ∠E = 63°
∴ ∠C = 180° – (∠A + ∠B) = 180° – (47° + 63°) = 70°
∴ Given statement is true.


Let ∆ABC ~ ∆DEF and their areas be respectively 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

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જવાબ :


ABC is an isosceles triangle right-angled at C. Prove that AB2 = 2AC2.

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જવાબ : ∆ABC is right-angled at C.
∴ AB2 = AC2 + BC2 [By Pythagoras theorem]
⇒ AB2 = AC2 + AC2
[∵ AC = BC]
⇒ AB2 = 2AC2


Sides of triangle are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm

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જવાબ : (i) Let a = 7 cm, b = 24 cm and c = 25 cm.
Here, largest side, c = 25 cm
We have, a2 + b2 = (7)2 + (24)2 = 49 + 576 = 625 = c2 [∵c = 25]
So, the triangle is a right triangle.
Hence, c is the hypotenuse of right triangle. (ii) Let a = 3 cm, b = 8 cm and c = 6 cm
Here, largest side, b = 8 cm
We have, a2 + c2 = (3)2 + (6)2 = 9 + 36 = 45 ≠ b2
So, the triangle is not a right triangle.


If triangle ABC is similar to triangle DEF such that 2AB = DE and BC = 8 cm. Then find the length of EF.

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જવાબ : ∆ABC ~ ∆DEF (Given)


If the ratio of the perimeter of two similar triangles is 4 : 25, then find the ratio of the areas of the similar triangles.

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જવાબ : ∵ Ratio of perimeter of 2 ∆’s = 4 : 25
∵ Ratio of corresponding sides of the two ∆’s = 4 : 25
Now, the ratio of area of 2 ∆’s = Ratio of square of its corresponding sides.
= (4)2(25)2 = 16625


In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2, then find ∠C.

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જવાબ :
AB2 = 2AC2 (Given)
AB2 = AC2 + AC2
AB2 = AC2 + BC2 (∵ AC = BC)
Hence AB is the hypotenuse and ∆ABC is a right angle A.
So, ∠C = 90°


The length of the diagonals of a rhombus are 16 cm and 12 cm. Find the length of side of the rhombus.

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જવાબ : ∵ The diagonals of rhombus bisect each other at 90°.
∴ In the right angle ∆BOC
BO = 8 cm
CO = 6 cm
∴ By Pythagoras Theorem
BC2 = BO2 + CO2 = 64 + 36
BC2 = 100
BC = 10 cm


A man goes 24 m towards West and then 10 m towards North. How far is he from the starting point?

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જવાબ : By Pythagoras Theorem
AC2 = AB2 + BC2 = (24)2 + (10)2
AC2 = 676
AC = 26 m
∴ The man is 26 m away from the starting point.


∆ABC ~ ∆DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of ∆DEF is 25 cm, what is the perimeter of ∆ABC?

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જવાબ : Since ∆ABC ~ ∆DEF.

 


. ∆ABC ~ ∆PQR; if area of ∆ABC = 81 cm2, area of ∆PQR = 169 cm2 and AC = 7.2 cm, find the length of PR.

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જવાબ : Since ∆ABC ~ ∆PQR


In Fig. 7.10, DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.

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જવાબ : In ∆ABC, we have
DE || BC,
∴ ADDB = AEEC [By Basic Proportionality Theorem]
⇒ xx−2 = x+2x−1
⇒ x(x – 1) = (x – 2) (x + 2)
⇒ x2 – x = x2 – 4
⇒ x = 4


E and F are points on the sides PQ and PR respectively of a ∆PQR. Show that EF ||QR if PQ = 1.28 cm, PR= 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

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જવાબ : We have, PQ = 1.28 cm, PR = 2.56 cm
PE = 0.18 cm, PF = 0.36 cm
Now, EQ = PQ-PE = 1.28 – 0.18 = 1.10 cm and
FR = PR – PF = 2.56 – 0.36 = 2.20 cm Therefore, EF || QR [By the converse of Basic Proportionality Theorem]


In Fig. 7.10, DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.

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જવાબ : In ∆ABC, we have
DE || BC,
∴ ADDB = AEEC [By Basic Proportionality Theorem]
⇒ xx−2 = x+2x−1
⇒ x(x – 1) = (x – 2) (x + 2)
⇒ x2 – x = x2 – 4
⇒ x = 4


E and F are points on the sides PQ and PR respectively of a ∆PQR. Show that EF ||QR if PQ = 1.28 cm, PR= 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

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જવાબ :


In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

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જવાબ :


R and S are points on the sides DE and EF respectively of a ADEF such that ER = 5 cm, RD = 2.5 cm, SE = 1.5 cm and FS = 3.5 cm. Find whether RS || DF or not.

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જવાબ :


In the figure, PQR and QST respectively. Prove that QR X QS= QP X QT

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જવાબ :


In ADEW, AB || EW. If AD = 4 cm, DE = 12 cm and DW = 24 cm, then find the value of DB.

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જવાબ :


A ladder is placed against a wall such that its foot is at a distance of 5 m from the wall and its top reaches a window 5/3 m above the ground. Find the length of the ladder

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જવાબ :


Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Using the above, prove the following:
If the areas of two similar triangles are equal, then prove that the triangles are congruent.

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જવાબ :


In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite the first side is a right angle.
Using the above, do the following:
In an isosceles triangle PQR, PQ = QR and PR2 = 2PQ2. . Prove that ZQ is a right angle.

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In fig. ∠M =∠N = 46°, express JC in terms of a, b, and c, where a, and c are lengths of LM, MN, and NK respectively.

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જવાબ :


If the areas of two similar triangles are in ratio 25 : 64, write the ratio of their corresponding sides.

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In a ∆ABC, DE || BC. If DE = – BC and area of ∆ABC = 81 cm2, find the area of ∆ADE.

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In figure,∆ABD is a right triangle, right angled at A and AC⊥BD. Prove that AB2 = BC.BD.

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In the figure, AABC is right-angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.

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Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided is the same ratio.
Using the above result, do the following:
In figure, DE|| BC and BD = CE. Prove that ∆ABC is an isosceles triangle.

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In the given figure PQ || BA; PR || CA. If PD = 12 cm. Find BD X CD.

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જવાબ :


In the given figure, if AB || DC, find the value of x.

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જવાબ :


Prove that in a right angle triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides. ‘
Using the above, do the following:
Prove that in a ∆ABC, if AD is perpendicular to BC, then AB2 + CD2 = AC2 + BD2.

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જવાબ :


AABC is an isosceles triangle in which AC = BC. If AB2  =   2AC2 then, prove that ∆ABC is right triangle.

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જવાબ :


In ∆ABC, AD is the median to BC and in ∆PQR, PM is the median to QR.

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જવાબ :


Prove that in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

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જવાબ :


In the figure, ABC is a right triangle, right-angled at B. AD and CF are two medians drawn from A and C respectively. If AC = 5 cm and AD = 3√5/2cm. Find the length of CE.

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In the given figure, ABCD is a rectangle. P is the mid-point of DC. If QB = 7 m, AD = 9 cm and DC = 24 cm, then prove that ∠APQ = 90°

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જવાબ :


In ∆ABC, ZB = 90°, BD ⊥ AC, ar (∆ABC) = A and BC = a, then prove that

 

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જવાબ :


Prove that the sum of square of the sides of a rhombus is equal to the sum of squares of its diagonals.

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જવાબ :


Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then other two sides are divided in the same ratio.

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જવાબ :


In ∆ABC, AX⊥ BC and Y is the middle point of BC. Then prove that,

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જવાબ :


Equiangular triangles are drawn on sides of right angled triangle in which perpendicular is double of its base. Show that area of triangle on the hypotenuse is the sum of areas of the other two triangles?

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જવાબ :


In the given figure, ABC is a triangle, right angled at B and BD⊥AC. If AD = 4 cm and CD = 5 cm, find BD and AB.

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જવાબ :


In the given figure, if LM || CB and LN|| CD, prove that AM X AD = AB X AN.

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જવાબ :


In the figure, PQR and QST respectively. Prove that QR X QS= QP X QT

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જવાબ :


State and prove Converse of Pythagoras’ Theorem.

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જવાબ :


From the airport two aeroplanes start at the same time. If the speed of first aeroplane due North is 500 km/h and that of the other due East is 650 km/h, then find the distance between the two aeroplanes after 2 hours.

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જવાબ :


 In figure, AB || PQ || CD, AB = x units, CD =y units and PQ = z units, prove that

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જવાબ :


Prove that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Using the above, do the following:
In a trapezium ABCD, AC and BD intersecting at O, AB|| DC and AB = 2CD, if area of ∆AOB = 84 cm2, find the area of ∆COD

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જવાબ :


In the figure, AD ⊥ BC and BD = 1/3 CD.
Prove that 2CA2 = 2AB2 + BC2.

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જવાબ :


In figure, S and T trisect the side QR of a right triangle PQR. Prove that 8PT2 = 3PR2 + 5PS2.

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જવાબ :


State and prove Converse of Pythagoras’ Theorem.

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જવાબ :


ABC is right-angled at C. If p is the length of the perpendicular from C to AB and a, b, c are the lengths of the sides opposite ∠A,∠B, ∠C respectively then prove that 

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જવાબ :


In the figure, ABCD is a parallelogram and E divides BC in the ratio 1: 3. DB and AE intersect at F. Show that DF = 4FB and AF = 4FE

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જવાબ :


ABC is a triangle in which AB = AC and D is a point on AC such that BC2 = AC × CD. Prove that BD = BC.
 

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જવાબ : Given: ∆ABC in which AB = AC and D is a point on the side AC such that BC2 = AC × CD
To prove: BD = BC
Construction: Join BD

Proof: We have,


 In a triangle, if the square on one side is equal to the sum of the squares on the other two sides, prove that the angle opposite to the first side is a right angle. Use the above theorem to find the measure of ∠PKR in Fig. 7.51.

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જવાબ : Given: A triangle ABC in which AC2 = AB2 + BC2
To Prove: ∠B = 90°.
Construction: We construct a ∆PQR right-angled at Q such that PQ = AB and QR = BC
Proof: Now, from ∆PQR, we have,


Prove that, in a ∆ABC if AD is perpendicular to BC, then AB2 + CD2 = AC2 + BD2.

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જવાબ :
Given: A right triangle ABC right-angled at B.
To Prove: AC2 = AB2 + BC2
Construction: Draw BD ⊥ AC
Proof: In ∆ADB and ∆ABC
∠A = ∠A (Common)
∠ADB = ∠ABC (Both 90°)
∴ ∆ADB ~ ∆ABC (AA similarity criterion)

Adding (i) and (ii), we get
AD. AC + CD . AC = AB2 + BC2
or, AC (AD + CD) = AB2 + BC2
or, AC . AC = AB2 + BC2
or, AC2 = AB2 +BC2

Second Part:
In Fig. 7.50, As AD ⊥ BC
Therefore, ∠ADB = ∠ADC = 90°
By Pythagoras Theorem, we have
AB2 = AD2 + BD2 …..(i)
AC2 = AD2 + DC2 …..(ii)
Subtracting (ii) from (i)
AB2 – AC2 = AD2 + BD2 – (AD2 + DC2)
⇒ AB2 – AC2 = BD2 – DC2 = AB2 + DC2 = BD2 + AC2


In Fig. 7.37, ABCD is a trapezium with AB || DC. If ∆AED is similar to ΔBEC, prove that AD = BC.

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જવાબ : In ∆EDC and ∆EBA we have
∠1 = ∠2 [Alternate angles]
∠3 = ∠4 [Alternate angles]
∠CED = ∠AEB [Vertically opposite angles]
∴ ∆EDC ~ ∆EBA [By AA criterion of similarity]


If AD and PM are medians of triangles ABC and PQR respectively, where ∆ABC ~ ∆PQR, prove that ABPQ = ADPM

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જવાબ :


ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AOBO = CODO.
 

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જવાબ : Given: ABCD is a trapezium, in which AB || DC and its diagonals intersect each other at point O.

 


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