Draw a right angled ∆PQR which is right angled at Q, in with the sides PQ and QR are of lengths 4 cm and 3 cm respectively. Then, construct another triangle whose sides are 3/ 5 times the corresponding sides of the given Triangle.
(Class 10 Maths Sample Question Paper)
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Answered by
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FIGURE IS IN THE ATTACHMENT.
SOLUTION:
Given: A ∆PQR in which ∠Q= 90°, PQ =4 cm and QR= 3 cm.
Steps of construction:
1.Draw line segment PQ =4 cm.
2. At Q make ∠PQY= 90°.
3. Take Q as centre and radius 3cm draw an Arc cutting QY at R.
4 Join PR to obtain the ∆PQR.
5. Through Q, construct an acute angle ∠PQX(<90°) opposite to the verteR.
6. Mark 5 points Q1, Q2, Q3, Q4, Q5 in QX, such that QQ1= Q1Q2= Q2Q3= Q3Q4=Q4Q5.
7.Join Q5P.
8.Through Q3 draw Q3P1||Q5P intersecting QP as P1.
9. Now through PQ ,draw P1R1||PR intersecting QR at R1.
Hence, ∆P1QR1 is the required ∆.
HOPE THIS WILL HELP YOU....
SOLUTION:
Given: A ∆PQR in which ∠Q= 90°, PQ =4 cm and QR= 3 cm.
Steps of construction:
1.Draw line segment PQ =4 cm.
2. At Q make ∠PQY= 90°.
3. Take Q as centre and radius 3cm draw an Arc cutting QY at R.
4 Join PR to obtain the ∆PQR.
5. Through Q, construct an acute angle ∠PQX(<90°) opposite to the verteR.
6. Mark 5 points Q1, Q2, Q3, Q4, Q5 in QX, such that QQ1= Q1Q2= Q2Q3= Q3Q4=Q4Q5.
7.Join Q5P.
8.Through Q3 draw Q3P1||Q5P intersecting QP as P1.
9. Now through PQ ,draw P1R1||PR intersecting QR at R1.
Hence, ∆P1QR1 is the required ∆.
HOPE THIS WILL HELP YOU....
Attachments:
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Answered by
1
Answer:
Given
PQ =4 cm
QR= 3cm
<Q=90
Step-by-step explanation:
Attachments:
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