NCERT Class 10 Chapter 7 Triangles Official CBSE Board Sample Problems Long Answer

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Question

In , if . then prove that . Also, if , and , then find DE.

Solution

Triangles

Triangles

Given: . i.e.

To prove:

Proof: In and

[Given]

[Common]

So, [By AA similarity]

Hence, .

Question

In, from A and B altitudes AD and BE are drawn. Prove that. Is and

Solution

Triangles

Triangles

In and ,

(Each )

(Common)

So, (By AA similarity)

is not similar to.

and is not similar to .

Question

State and prove Converse of Pythagoras’ Theorem.

Solution

Triangles

Triangles

Statement: 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 to the first side is right angle.

Given: In

To prove:

Construction: Construct a , such that and and

Proof: In , (Given)

(By Pythagoras)

(Using and by construction) ..(i)

but (Given) ...(ii)

Equating (1) and (ii)

Now, in and

(By construction)

(by construction)

(Proved)

(By SSS)

Hence

This shows that, MBC is an right-angled triangle.

Question

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.

Solution

Triangles

Triangles

Given: A triangle ABC in which

and DE interacts AB in D and AC in E.

To prove-

Construction: Join BE, CD and draw and

Proof: is perpendicular to AB.

is height of triangles ADE and DBE

Now, .. (i)

Similarly. So is height of and .

..(ii)

Ut, and arc on same basc DE and between same parallels DE and BC. ar(DBE) •

Multiplying both side by

Hus, [from (i) and (ii)]

Question

In and Y is middle point of BC. Then prove that,

(i)

(ii)

Solution

Given: A in which and Y is mid-point of BC.

To prove: (i)

(ii)

Proof: (i) In .

[BY Pythagoras Theorem]

(Y is midpoint of BC)

[In]

Hence proved.

(ii) In ,

[By Pythagoras Theorem]

: In Y is mid-point of BCI

(... In ]

Hence proved

Question

Prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Solution

Triangles

Triangles

Given: Two triangles ABC and DEF such that .

To prove:

Construction: Draw and

Proof: In and,

[BY Construction]

:. By AA criterion of similarity,

So,

...(II) (Ratio of corresponding sides of similar triangles arc equal]

From (I) and (ii), we get

...(iii)

Now,

(From (iii)]

And

Hence,

Hence proved

Question

O is any point inside a rectangle ABCD. Prove that:

.

Solution

Rectangle

Rectangle

Through O, draw so that P lies on

AB and Q lies on DC.

Now,

Therefore, and ( and ( )

So, and

Therefore, BPQC and APQD are both rectangles.

Now, from ,

…(1)

Similarly, Irvin

…(2)

From , we have

… (3)

and from . we have

…. (4)

Adding (1) and (2),

(As BP=CQ and DQ=AP)

[From (3) and (4)]

Question

In the given figure, line segment XY is parallel to side AC of triangle ABC and it divides the Triangle into two parts of equal areas Find the ratio .

Triangle

Triangle

Solution

In figure, the line segment XY is parallel to side AC of ABC and it divides the triangle into two parts of equal areas. Find the ratio

Given:

XY is parallel to AC i.e.

To find:

Proof:

Ln ,

(Common)

(Since XV I AC, corresponding angles are equal)

(AA similarity)

Triangle

Triangle

Now,

We know that in similar triangles,

Ratio of area of triangle is equal to ratio of square of corresponding sides

(As Area of =)

Question

AD and PM are medians of triangles ABC and PQR respectively, where . Prove that

Triangle

Triangle

Solution

Given: and

AD is the median of

PM is the median of

.

To Prove:-

Proof:

Since AD is the median

Similarly, PM is the median

Now,

Triangle

Triangle

.

(Corresponding sides of similar triangle ore proportional)

So,

(Since AD & PM are medians)

… (1)

PQ QM

Also, since.

(Corresponding angles of similar triangles ate equal)(2)

Now,

Triangle

Triangle

In

(From (2)

(From (1)

Hence by SAS similarly

Since corresponding sides of similar triangles are proportional

Hence proved

Question

The perpendicular drawn from A on side BC of a intersects BC at D such that Prove that

Solution

Triangle

Triangle

Given: with

Also

To prove:

Proof:

Let

(As DB=3CD given) =3x

and

Equating the two values of we get:

Question

D, E and F are respectively the mid points of the sides AB. BC and CA of triangle ABC respectively. Find the ratio of areas of triangle DEF and triangle ABC.

Solution

Question

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution

Let ABCD be a square and each side be a units

Let

and

be the equilateral triangles described on the side BC and diagonal AC respectively

(Pythagoras Theorem)

(All equilateral triangles are similar by AAA)

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