Here the concept or Areas of the Triangle has been used. We see that we are given two sub parts in the question where we have to find the length of a side. We are also given that there is one altitude and we have to find the other side. When we construct the triangle according to given notations, then we see that the side which we have to find that is QR is the base of triangle. Using this concept, we can get our answer.
Let'sdoit!!
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★FormulaUsed:-
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★Solution:-
a.) ar(∆PQR) = 180 cm², PS = 15 cm
Given,
» Area of ∆PQR = 180cm²
» Height of ∆PQR = Altitude = PS = 15cm
From figure we see that,
Baseof the triangle ∆PQR = QR
We know that,
By applying values, we get
By cross multiplication, we get
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b.) ar(∆PQR) = 111 cm², PS = 12 cm
Given,
» Area of ∆PQR = 111cm²
» Heightof ∆PQR = Altitude = PS= 12cm
From figure we see that,
Baseof ∆PQR = QR
We know that,
By applying values, we get
By cross multiplication, we get
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★Moretoknow:-
• Formulas::
•PropertiesofTriangle::
Its is made up of threesides.
A triangle whose all threesidesareequalis known as EquilateralTriangle.
A triangle whose two sides are equalis known as IsocelesTriangle.
A triangle whose allthreesidesareofdifferentmeasures, is known as ScaleneTriangle.
Answers & Comments
Here the concept or Areas of the Triangle has been used. We see that we are given two sub parts in the question where we have to find the length of a side. We are also given that there is one altitude and we have to find the other side. When we construct the triangle according to given notations, then we see that the side which we have to find that is QR is the base of triangle. Using this concept, we can get our answer.
Let's do it !!
_____________________________________________
★ Formula Used :-
_____________________________________________
★ Solution :-
a.) ar(∆PQR) = 180 cm², PS = 15 cm
Given,
» Area of ∆PQR = 180 cm²
» Height of ∆PQR = Altitude = PS = 15 cm
From figure we see that,
We know that,
By applying values, we get
By cross multiplication, we get
_____________________________________________
b.) ar(∆PQR) = 111 cm², PS = 12 cm
Given,
» Area of ∆PQR = 111 cm²
» Height of ∆PQR = Altitude = PS = 12 cm
From figure we see that,
We know that,
By applying values, we get
By cross multiplication, we get
_____________________________________________
★ More to know :-
• Formulas ::
• Properties of Triangle ::
Verified answer
Answer:
(a) Area of ∆PAR= 1/2*b*h=1/2*QR*PS
180=1/2*15*QR
QR=180*2/15=24cm
(b) Area of ∆PAR= 1/2*b*h=1/2*QR*PS
111=1/2*12*QR
QR=111*2/12=18.5cm
I hope my answer helps you ☺☺