Question

a) Draw triangle PQR with PQ = 8cm, PR = 6cm, angle P= 30
b) Draw a right triangle of the same area as that of triangle PQR.​

Answer 1

Given

PQ=8cm and QR=6cm

∠PQR=90°

To find

Length of PR

Solution

We can simply solve this mathematical problem by using the following mathematical process.

Now,  

Using Pythagoras Theorem

$PR^{2}$ = $PQ^{2}$ + $QR^{2}$

Substituting the values

$PR^{2}$ = $8^{2}$ + $6^{2}$

By further calculation

$PR^{2}$ = 64 + 36 = 100

PR=√100

So we get

PR = 10cm

Hence the length of PR is 10cm

Answer 2

Drawn triangle PQR where PQ = 8 cm , PR = 6 cm , ∠P = 30°

Drawn a right triangle PQR' of the same area as that of triangle PQR.​

Given:

• PQ = 8 cm
• PR = 6 cm
• ∠P = 30°

To Find:

• Draw triangle PQR
• Draw a right triangle of the same area as that of triangle PQR.​

Solution:

Draw triangle PQR

Step 1:  Draw  a line segment PQ = 8 cm using ruler

Step 2: Draw an ray PX at an angle of 30° at PQ using protractor

Step 3: Using compass width = 6 cm and taking P as center draw an arc intersecting PX at R

Step 4: Join QR

ΔPQR is constructed

Draw a right triangle of the same area as that of triangle PQR.​

Step 1:  Draw a ray QY at right angle  on PQ using protractor/set square

Step 2: Draw a line parallel to PQ using setsquare passing through R and intersecting QY at R'

Step 3: Join PR'

ΔPQR' is constructed which is a right triangle of the same area as  ΔPQR

(∵ Base and height are same)