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 April 16, 2024 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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