ELAMPLE 1 Construct a triangle PQR whose perimeter is equal to 14 cm,
angle P = 45°
and
angle Q = 60°
Answers
Step-by-step explanation:
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< b > < h4 > Here is your answer < /h4 > ; Please refer to the given attachment ; < /b ><b><h4>Hereisyouranswer</h4>;Pleaserefertothegivenattachment;</b>
Given,
Perimeter is 14cm.
<P = 45.
<Q = 60.
< b > Steps; < /b ><b>Steps;</b>
1) Draw the perimeter XY 14cm.
2) Construct <AXP = 45 and <QYB = 60.
3) Draw the angle bisectors of both <AXP and <QYB, which both meet at R.
4) Now, draw the perpendicular bisectors of RX and RY formed which intersect at P and Q respectively.
5) Join R with P and Q.
6) PQR is the required triangle.
< b > Verification; < /b ><b>Verification;</b>
PR = 4.5Cm, QR = 4Cm, and PQ = 5.5Cm
Adding PR, QR, And PQ, we get,
Perimeter = 14Cm.
Also on measuring,
<P = 45.
<Q = 60.
< b > Proof < /b ><b>Proof</b>
P lies on the perpendicular bisector of XR.
Thus, XP = PR.
Q lies on the perpendicular bisector of YR.
Thus, QY = QR.
Therefore,
XP + PQ + QY = perimeter.
=> PR + PQ + QR = 14Cm.
Also,
XP = PR
=> <XRP = <PXR
<RPQ = <PXR + <PRX as exterior angle of triangle XRP.
=> <RPQ = 2<PXR
= 2 * \frac{45}{2}2∗
2
45
= 45 degree.
Same as above,
QY = QR
=> <QRY = <QYR
<PQR = <QRY + <QYR as exterior angle of triangle QRY.
=> <PQR = 2<QYR
= 2 * \frac{60}{2}2∗
2
60
= 60 degree.
< b > < h1 > < marquee > < font color =green > I hope it helps you #need more brainliests marks
