NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.4

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.4 are provided here to help students understand and apply the Pythagoras Theorem and its converse. These solutions are prepared strictly according to the latest CBSE syllabus and are explained in a simple, step-by-step manner for better clarity.

Exercise 6.4 mainly focuses on:

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.4

Triangles

Medium

Q.

$\text{In the following figure, }\Delta \text{ODC}\sim \Delta \text{OBA, }\angle \text{BOC}=125°\text{and }\angle \text{CDO}=70°.\text{ Find }\angle \text{DOC, }\angle \text{DCO and }\angle \text{OAB}\text{.}$

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Triangles

Medium

Q.

$\text{Let }\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }{\text{DEF and their areas be, respectively, 64 cm}}^{\text{2}}{\text{ and 121 cm}}^{\text{2}}\text{. If EF}=\text{15.4 cm, find BC.}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the}\\ \text{equilateral triangle described on one of its diagonals.}\end{array}$

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Triangles

Medium

Q.

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

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Triangles

Medium

Q.

$\begin{array}{l}\text{D, E and F are respectively the mid-points of sides AB, BC and CA of }\mathrm{\Delta }\text{ABC. Find the ratio of the areas}\\ \text{of }\mathrm{\Delta }\text{DEF and }\mathrm{\Delta }\text{ABC.}\end{array}$

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

In the following figure, ABC and DBC are two triangles on the same base BC.
If AD intersects BC at O, show that ar(ΔABC) (ΔDBC) = AO DO.

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Triangles

Medium

Q.

Diagonals of a trapezium ABCD with AB ║ DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

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Triangles

Medium

Q.

$\begin{array}{l}\text{If AD and PM are medians of triangles ABC and PQR, respectively where }\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR, prove that}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}.\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{Diagonals AC and BD of a trapezium ABCD with AB}\parallel \text{DC intersect each other at the point O. Using}\\ \text{a similarity criterion for two triangles, show that }\frac{\text{OA}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}.\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower}\\ \text{casts a shadow 28 m long. Find the height of the tower.}\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{Sides AB and AC and median AD of a triangle ABC }\\ \text{are respectively proportional to sides PQ and PR and }\\ \text{median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR}\end{array}$

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Triangles

Medium

Q.

$\text{D is a point on the side BC of a triangle ABC such that }\angle \text{ADC}=\angle {\text{BAC. Show that CA}}^{\text{2}}=\text{CB}·\text{CD.}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{CD and GH are respectively the bisectors of }\mathrm{\Delta }\text{ACB and }\mathrm{\Delta }\text{EGF such that D and H lie on sides AB and FE}\\ \text{of }\mathrm{\Delta }\text{ABC and }\mathrm{\Delta }\text{EFG respectively. If }\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{FEG, show that:}\\ \text{(i) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{CD}}{\text{GH}}=\frac{\text{AC}}{\text{FG}}\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCB}~\mathrm{\Delta }\text{HGE}\\ \text{(iii)\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCA}~\mathrm{\Delta }\text{HGF}\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD}\\ \text{at F. Show that }\mathrm{\Delta }\text{ABE}~\mathrm{\Delta }\text{CFB.}\end{array}$

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Triangles

Medium

Q.

$\text{S and T are points on sides PR and QR of }\mathrm{\Delta }\text{PQR such that }\angle \text{P}=\angle \text{RTS. Show that }\mathrm{\Delta }\text{RPQ}~\mathrm{\Delta }\text{RTS.}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles}\\ \text{ABC and BDE is}\\ \text{(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4}\end{array}$

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• Application of Pythagoras Theorem

• Verification of whether a triangle is right-angled

• Finding unknown sides of a triangle

• Using the converse of Pythagoras Theorem

Students learn that in a right-angled triangle:

$$(\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2$$

(Hypotenuse)2=(Base)2+(Perpendicular)2

This exercise helps students strengthen their calculation skills and logical reasoning, which are important for board examinations.

These solutions are designed to help students practice efficiently and score better in exams.

Q1. Let ΔABC ~ ΔDEF and their areas be 64 cm² and 121 cm² respectively.
If EF = 15.4 cm, find BC.

🔎 Reasoning:

Ratio of the areas of two similar triangles = square of the ratio of their corresponding sides.

✅ Solution:

Q2. Diagonals of trapezium ABCD (AB || DC) intersect at O.
If AB = 2CD, find ratio of areas of ΔAOB and ΔCOD.

🔎 Reasoning:

Triangles AOB and COD are similar (AA criterion).
Area ratio equals square of side ratio.

✅ Solution:

Given AB = 2CD,

Q3. In Fig. 6.44, ABC and DBC are triangles on same base BC.
If AD intersects BC at O, show that:

🔎 Reasoning:

Use AA similarity between ΔAOM and ΔDON.

✅ Result:

Q4. If areas of two similar triangles are equal, prove they are congruent.

🔎 Reasoning:

Area ratio = square of side ratio.

✅ Solution:

So,

Thus,

AB = DE
BC = EF
CA = FD

By SSS congruency,

Q5. D, E and F are midpoints of sides AB, BC and CA of ΔABC.
Find ratio of areas of ΔDEF and ΔABC.

🔎 Reasoning:

Midpoint theorem: DE = ½ BC
Triangles are similar.

✅ Solution:

Q6. Prove that ratio of areas of two similar triangles equals square of ratio of their corresponding medians.

✅ Result:

Q7. Area of equilateral triangle on side of square equals half area of equilateral triangle on diagonal.

🔎 Reasoning:

Diagonal of square = side × √2

Thus,

Area on side = ½ × Area on diagonal

Q8. ABC and BDE are equilateral triangles. D is midpoint of BC.
Find ratio of areas.

Since BD = ½ BC,

✅ Answer:

4 : 1 (Option C)

Q9. Sides of two similar triangles are in ratio 4 : 9.
Find ratio of areas.

✅ Answer:

16 : 81 (Option D)

FAQs: Class 10 Maths Chapter 6 – Triangles Exercise 6.4

Q1. What is the main concept of Exercise 6.4?

Answer:
Exercise 6.4 focuses on the Pythagoras Theorem and its converse to solve problems related to right-angled triangles.

Q2. What is Pythagoras Theorem?

Answer:
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Q3. What is the converse of Pythagoras Theorem?

Answer:
If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right-angled triangle.

Q4. Why is Exercise 6.4 important for exams?

Answer:
Questions based on Pythagoras Theorem are commonly asked in CBSE board exams, especially for checking whether a triangle is right-angled.

Q5. How do NCERT Solutions help in this exercise?

Answer:
NCERT Solutions provide clear explanations, accurate calculations, and step-by-step methods that help students understand the concept thoroughly and avoid mistakes in exams.