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Is the following Byjus Class 7 Maths Chapter 2 Answer statement true? Solution: False Two quadrilaterals are similar if their corresponding angles are equal and corresponding sides must also be proportional.

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? Solution: True Here, the corresponding two sides and the perimeters of two triangles are proportional, then the third side of both triangles will also in proportion. If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles will be similar?

Is it true to say that if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Solution: False Because, according to SAS similarity criterion, if one angle of a triangle is equal to an angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Here, one angle and two sides of two triangles are equal but these sides not including equal angle, so given statement is not correct. AP Hence proved. Find the altitude of an equilateral triangle of side 8 cm. Solution: Let ABC be an equilateral triangle of side 8 cm i.

Then, D is the mid-point of BC. Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm 2 , find the area of the larger triangle.

If PN. Areas of two similar triangles are 36 cm 2 and cm 2. If the length of a side of the larger triangle is 20 cm, find the length of the corresponding side of the smaller triangle. A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

Hence, the height of the point on the wall where the top of the ladder reaches is 8 m. Foot of a 10 m Ch 4 Maths Class 10 Ex 4.3 Vol.1 long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

The top of the ladder reached to A and distance of ladder from the base of the wall BC is 6 m. Find the lengths of the remaining sides of the triangles. Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio. To prove : DE divides the two sides in the same ratio.

Also AB PS. Hence proved. A Ncert Class 10 Maths Ch 6 Ex 6.3 Vo 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1. You also need to give the correct notation for describing similar triangles. In the second question of Class 10 Maths Chapter 6 Exercise 6. You need to apply the theorem for alternate interior angles and vertically opposite angles to solve this problem.

In this question, a triangle is given, which contains two triangles inside of it. By applying the theorem that if the sides of a triangle are proportional, then their corresponding angles are equal, you can prove that the two interior triangles are similar. This is a straightforward question where two angles of two triangles are the same AA criterion hence students can prove that the given triangles are similar. This question of Class 10 Maths Exercise 6.

In this question of Triangles Class, 10 Exercise 6. You will be using the concept of vertically opposite angles and common angles to solve this question. This is again a simple question where two right triangles are given which have a common side, and you need to prove that both are similar triangles which can be easily proved with a common angle, corresponding sides of similar triangles, and alternate angle concept.

Class 10 Maths Exercise 6. In this problem, an isosceles triangle is given, and one of its sides is extended, and a perpendicular is dropped from the extended point to the opposite side of the triangle.

Students need to prove that the triangle formed by extension and the triangle formed by dropping a perpendicular to the base of the original triangle are similar. The properties of the isosceles triangle and alternate angle criterion are used to solve this problem.

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