Math, asked by bmchavan, 6 months ago

plz solve this question
lesson 4 std 10 ssc
construction ​

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Answers

Answered by amitsnh
0

Answer:

construction of a triangle when all sides are given is easy.

Regarding ∆ PQR

PQ = 3.2 cm

QR = 3.6 cm

PR = 7.2 cm

now we are in big trouble!!!!

PQ + QR = 3.2 + 3.6 = 6.8 < 7.2

PQ + QR < PR

this is violating the very basic condition for construction of a triangle. Hence, a triangle with these measurements cannot be made.

let us change PQ to 3.8 cm. (assuming you wrote 2 in place of 8 mistakenly)

our new dimensions are

PQ = 3.8 cm

QR = 3.6 cm

PR = 7.2 cm

now draw a horizontal line and mark QR = 3.6 cm on it.

from point Q make an arc of 3.8 cm radius using compass. Similarly, from point R, make an arc of 7.2 cm using compass such that it intersects the earlier made arc at some point. mark the point of intersection as P

Join PQ and PR and it is done!!!!!

Regarding ∆STU

given ∆PQR is similar to ∆STU and

PQ/ST = 5/4

ST = PQ*4/5

= 3.8*4/5

= 15.2/5

= 3.04 cm

similarly

PR/SU = QR/TU = 5/4

SU = PR*4/5

= 7.2*4/5

= 28.8/5

= 5.76 cm

TU = QR*4/5

= 3.6*4/5

= 14.4/5

= 2.88 cm

Now that we have measurement of all the sides of ∆STU, we can draw the triangle in same way as we did in case of ∆PQR.

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