Question

. Prove that the lengths of altitudes drawn to equal sides of an isosceles triangle
are also equal.
(i) ∠TRQ = ∠SQR?
(ii) If ∠TRQ = 30°, find the base angles of the ΔPQR.
(iii) Is ΔPQR an equilateral triangle?

Answer 1

Answer:

Step-by-step explanation:

Congruence of Triangles Class 7 Extra Questions Maths Chapter 7

June 12, 2019 by Sastry CBSE

Congruence of Triangles Class 7 Extra Questions Maths Chapter 7

Extra Questions for Class 7 Maths Chapter 7 Congruence of Triangles

Congruence of Triangles Class 7 Extra Questions Very Short Answer Type

Question 1.

In the given figure, name

(a) the side opposite to vertex A

(b) the vertex opposite A to side AB

(c) the angle opposite to side AC

(d) the angle made by the sides CB and CA.

Congruence of Triangles Class 7 Extra Questions Maths Chapter 7

Solution:

(a) The side opposite to vertex A is BC.

(b) The vertex opposite to side AB is C.

(c) The angle opposite to side AB is ∠ACB.

(d) The angle made by the sides CB and CA is ∠ACB.

Question 2.

Examine whether the given triangles are congruent or not.

Solution:

Here,

AB = DE = 3 cm

BC = DF = 3.5 cm

AC = EF = 4.5 cm

ΔABC = ΔEDF (By SSS rule)

Congruence of Triangles Class 7 Extra Questions Maths Chapter 7

Answer 2

That the lengths of altitudes drawn to equal sides of an isosceles triangle

are also equal.

(i) ∠TRQ = ∠SQR?

(ii) If ∠TRQ = 30°, find the base angles of the ΔPQR.

(iii) Is ΔPQR an equilateral triangle

To prove that the lengths of altitudes drawn to equal sides of an isosceles triangle are also equal, we can consider the following diagram:

                     P

                    /  \

                  /      \

            Q /_____\ R

Let PQ = PR = a be the equal sides of the triangle, and let h1 and h2 be the lengths of the altitudes drawn from P to QR and from Q to PR, respectively.

To prove that h1 = h2, we can use the fact that the area of a triangle is given by the formula:

Area = (1/2) * base * height

Let's consider the area of triangle PQR. We can see that the base of the triangle is a, and its height is h1 + h2. Therefore, we have:

Area(PQR) = (1/2) * a * (h1 + h2)

Now, let's consider the areas of the triangles PQR and QPR separately. We know that these areas are equal, since they have the same base (a) and height (h1 and h2, respectively). Therefore, we have:

Area(PQR) = Area(QPR)

Substituting the formula for the area of a triangle into this equation, we get:

(1/2) * a * (h1 + h2) = (1/2) * a * h2

Simplifying and solving for h2, we get:

h2 = (h1 + h2)/2

h1 = h2

Therefore, we have proved that the lengths of altitudes drawn to equal sides of an isosceles triangle are also equal.

(i) ∠TRQ = ∠SQR?

Since triangle PQR is isosceles, we know that angles QPR and QRP are equal. Therefore, we have:

∠TRQ = 180° - ∠QRP = 180° - ∠QPR = ∠SQR

(ii) If ∠TRQ = 30°, find the base angles of the ΔPQR.

Since ∠TRQ = ∠SQR = 30°, we have:

∠QPR = (180° - 2*∠TRQ)/2 = (180° - 60°)/2 = 60°

Therefore, the base angles of triangle PQR are both equal to 60°.

(iii) Is ΔPQR an equilateral triangle?

No, we cannot conclude that triangle PQR is equilateral based on the given information. We know that the base angles are both equal to 60°, but we do not have any information about the length of the base itself. An isosceles triangle can have two equal angles without having three equal sides.

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