Practice Set 1.1 Similarity Class 10th Mathematics Part 2 MHB Solution

Practice Set 1.1

1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is…
2. If figure 1.13 BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find a (deltaabc)/a (deltaadb)…
3. In adjoining figure 1.14 seg PS seg RQ , seg QT seg PR. If RQ = 6, PS = 6 and PR = 12,…
4. In adjoining figure, AP ⊥ BC, AD || BC, then find A(ΔABC) : A (ΔBCD) n…
5. In adjoining figure PQ BC, AD BC then find following ratios. (i) a (deltapqb)/a…

Practice Set 1.1

Question 1.

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.

Answer:

We know that area of triangle =  × Base× Height

⇒ Area (triangle 1) =  ×9× 5

= 

⇒ Area (triangle 2) =  ×10× 6

= 30

∴ the ratio of areas of these triangles will be = 

= 

= 

= 

Question 2.

If figure 1.13 BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find 

Answer:

Here,ΔABC and ΔADB has common Base.

∴ 

(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)

⇒ 

= 

= 

Question 3.

In adjoining figure 1.14 seg PS ⊥ seg RQ , seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.

Answer:

Considering, Area of (ΔPQR) with base QR

⇒ PS will be the Height

Now, consider the Area of (ΔPQR) with base PR

⇒ QT will be the Height

∵ , the triangle is the same


⇒ the area will be the same irrespective of the base taken.

And we know that area of triangle =  × Base× Height

⇒ ×QR×PS

= ×PR×QT

⇒ ×6×6


= ×12×QT

⇒ QT = 3

Question 4.

In adjoining figure, AP ⊥ BC, AD || BC, then find A(ΔABC) : A (ΔBCD)

Answer:

We can re-draw the fig.1.15(as shown above) where we add DO

which will be height of ΔBCD.

Now, 

(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)

⇒

⇒ 

(∵ the distance between the two parallel lines is always equal ⇒ AP = DO)

⇒ = 1:1

Question 5.

In adjoining figure PQ ⊥ BC, AD ⊥ BC then find following ratios.


(i) 

(ii) 

(iii) 

(iv) 

Answer:

We know that area of triangle =  × Base× Height

(i)

(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)

(ii)

(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)

(iii)

(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)

(iv)

=