Question 35 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at Nov. 1, 2019 by Teachoo

Draw a triangle ABC with side BC = 6.5 cm, ∠B = 30°, ∠A = 105°. Then construct another triangle whose sides are 3/4 times the corresponding sides of the triangle ABC.

Note
: This
is similar
to Ex 11.1, 6 of NCERT – Chapter 11 Class 10

Question 35 (OR 1st question) Draw a triangle ABC with side BC = 6.5 cm, ∠B = 30°, ∠A = 105°. Then construct another triangle whose sides are 3/4 times the corresponding sides of the triangle ABC.
Let’s first draw a rough diagram
To construct Δ ABC, we first need to find ∠ C
Finding ∠ C
In Δ ABC
∠A + ∠B + ∠C = 180°
105° + 30° + ∠C = 180°
135° + ∠C = 180°
∠C = 180° − 135°
∠C = 45°
Steps to draw Δ ABC
Draw base BC of side 6.5 cm
Draw ∠ B = 30°
Draw ∠ C = 45°
4. Let point A be the point where the two rays intersect
∴ Δ ABC is the required triangle
Now, we need to make a triangle which is 3/4 times its size
∴ Scale factor = 3/4 < 1
Steps of construction
Draw any ray BX making an acute angle with BC
on the side opposite to the vertex A.
Mark 4 (the greater of 4 and 3 in 3/4 ) points
𝐵_1, 𝐵_2, 𝐵_3,𝐵_4 on BX so that 〖𝐵𝐵〗_1=𝐵_1 𝐵_2=𝐵_2 𝐵_3=𝐵_3 𝐵_4
Join 𝐵_4C
and draw a line through 𝐵_3 (the 3rd point, 3 being smaller of 3 and 4 in 3/4) parallel to 𝐵_4 𝐶,
to intersect BC at C′.
Draw a line through C′ parallel to the line AC
to intersect AB at A′.
Thus, Δ A′BC′ is the required triangle

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.