જવાબ :

જવાબ : ∆ABC – ∆DEF …[Given

જવાબ :

જવાબ : Let BD = x cm

then BW = (24 – x) cm, AE = 12 – 4 = 8 cm

In ∆DEW, AB || EW

જવાબ : In ∆ABC, DE || BC …[Given

x(x + 5) = (x + 3)(x + 1)

x^{2} + 5x = x^{2} + 3x + x + 3

x^{2} + 5x – x^{2} – 3x – x = 3

∴ x = 3 cm

જવાબ : In ∆ADE and ∆ABC,

∠DAE = ∠BAC …Common

∠ADE – ∠ABC … [Corresponding angles

∆ADE – ∆ΑΒC …[AA corollary

જવાબ : Let YR = x

PQXQ=PRYR … [Thales’ theorem

જવાબ : Altitude of an equilateral ∆,

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

જવાબ :

જવાબ : BE = BC – EC = 10 – 2 = 8 cm

Let AF = x cm, then BF = (13 – x) cm

In ∆ABC, EF || AC … [Given

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.

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ : Here, 12^{2}+ 16^{2} = 144 + 256 = 400 ≠ 18^{2}

∴ The given triangle is not a right triangle.

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.

જવાબ : Since ∆ABC ~ ∆DEF

∴ ∠A = ∠D = 47°

∠B = ∠E = 63°

∴ ∠C = 180° – (∠A + ∠B) = 180° – (47° + 63°) = 70°

∴ Given statement is true.

જવાબ :

જવાબ : ∆ABC is right-angled at C.

∴ AB^{2} = AC^{2} + BC^{2} [By Pythagoras theorem]

⇒ AB^{2} = AC^{2} + AC^{2}

[∵ AC = BC]

⇒ AB^{2} = 2AC^{2}

(i) 7cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

જવાબ : (i) Let a = 7 cm, b = 24 cm and c = 25 cm.

Here, largest side, c = 25 cm

We have, a^{2} + b^{2} = (7)^{2} + (24)^{2} = 49 + 576 = 625 = c^{2} [∵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, a^{2} + c^{2} = (3)^{2} + (6)^{2} = 9 + 36 = 45 ≠ b^{2}

So, the triangle is not a right triangle.

જવાબ : ∆ABC ~ ∆DEF (Given)

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

જવાબ :

AB^{2} = 2AC^{2} (Given)

AB^{2} = AC^{2} + AC^{2}

AB^{2} = AC^{2} + BC^{2} (∵ AC = BC)

Hence AB is the hypotenuse and ∆ABC is a right angle A.

So, ∠C = 90°

જવાબ : ∵ The diagonals of rhombus bisect each other at 90°.

∴ In the right angle ∆BOC

BO = 8 cm

CO = 6 cm

∴ By Pythagoras Theorem

BC^{2} = BO^{2} + CO^{2} = 64 + 36

BC^{2} = 100

BC = 10 cm

જવાબ :
By Pythagoras Theorem

AC^{2} = AB^{2} + BC^{2} = (24)^{2} + (10)^{2}

AC^{2} = 676

AC = 26 m

∴ The man is 26 m away from the starting point.

જવાબ : Since ∆ABC ~ ∆DEF.

જવાબ : Since ∆ABC ~ ∆PQR

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)

⇒ x^{2} – x = x^{2} – 4

⇒ x = 4

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]

DE || BC,

∴ ADDB = AEEC [By Basic Proportionality Theorem]

⇒ xx−2 = x+2x−1

⇒ x(x – 1) = (x – 2) (x + 2)

⇒ x^{2} – x = x^{2} – 4

⇒ x = 4

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

Using the above, prove the following:

If the areas of two similar triangles are equal, then prove that the triangles are congruent.

જવાબ :

Using the above, do the following:

In an isosceles triangle PQR, PQ = QR and PR

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

Using the above result, do the following:

In figure, DE|| BC and BD = CE. Prove that ∆ABC is an isosceles triangle.

જવાબ :

જવાબ :

જવાબ :

Using the above, do the following:

Prove that in a ∆ABC, if AD is perpendicular to BC, then AB

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

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

જવાબ :

Prove that 2CA

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ : Given: ∆ABC in which AB = AC and D is a point on the side AC such that BC^{2} = AC × CD

To prove: BD = BC

Construction: Join BD

જવાબ : Given: A triangle ABC in which AC^{2} = AB^{2} + BC^{2}

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,

જવાબ :

Given: A right triangle ABC right-angled at B.

To Prove: AC^{2} = AB^{2} + BC^{2}

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 = AB^{2} + BC^{2}

or, AC (AD + CD) = AB^{2} + BC^{2}

or, AC . AC = AB^{2} + BC^{2}

or, AC^{2} = AB^{2} +BC^{2}

In Fig. 7.50, As AD ⊥ BC

Therefore, ∠ADB = ∠ADC = 90°

By Pythagoras Theorem, we have

AB

AC

Subtracting (ii) from (i)

AB

⇒ AB

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]

જવાબ :

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

gseb maths textbook std 10

- Math Book for GSEB ધોરણ ૧૦
- Chemistry Book for GSEB ધોરણ ૧૦
- Biology Book for GSEB ધોરણ ૧૦
- Physics Book for GSEB ધોરણ ૧૦
- History Book for GSEB ધોરણ ૧૦
- Geography Book for GSEB ધોરણ ૧૦
- Economics Book for GSEB ધોરણ ૧૦
- Political Science Book for GSEB ધોરણ ૧૦

The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.