ML Aggarwal Construction of Quadrilaterals Exe-14.1 Class 8 ICSE Ch-14 Maths Solutions
ML Aggarwal Construction of Quadrilaterals Exe-14.1 Class 8 ICSE Ch-14 Maths Solutions. We Provide Step by Step Answer of Exe-14.1 Questions for Construction of Quadrilaterals as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.
ML Aggarwal Construction of Quadrilaterals Exe-14.1 Class 8 ICSE Maths Solutions
| Board | ICSE |
| Publications | Avichal Publishig Company (APC) |
| Subject | Maths |
| Class | 8th |
| Chapter-14 | Construction of Quadrilaterals |
| Writer | ML Aggarwal |
| Book Name | Understanding |
| Topics | Solution of Exe-14.1 Questions |
| Edition | 2023-2024 |
Construction of Quadrilaterals Exe-14.1
ML Aggarwal Class 8 ICSE Maths Solutions
Page-254
Question 1. Construct a quadrilateral PQRS where PQ = 4.5 cm, QR = 6 cm, RS = 5.5 cm, PS = 5 cm and PR = 6.5 cm.
Answer:
Steps of constructions :
(i) Draw a line segment PR = 6.5 cm.
(ii) With centre P and radius 4.5 cm and with centre R and radius 6 cm draw arcs intersecting each other at Q.
(iii) Join PQ and QR.
(iv) Similarly with centre P and radius 5 cm and with centre R and radius 5.5 cm, draw arcs intersecting each other at S.
(v) Join PS and SR. PQRS is the required quadrilateral.
Question 2. Construct a quadrilateral ABCD in which AB = 3·5 cm, BC = 5 cm, CD = 5·6 cm, DA = 4 cm and BD = 5·4 cm
Answer:
Steps of construction :
(i) Draw AB = 3·5 cm.
(ii) With A as centre and radius = 4 cm,
draw an arc with B as centre and radius = 5·4 cm
draw an arc to meet the previous arc at D. Join AD andBD.
(iii) With B as centre and radius = 5 cm,
draw an arc With D as centre and radius = 5·6 cm,
draw an arc to meet the previous arc at C.
(iv) Join BC and CD, then ABCD is the required quadrilateral.
Question 3. Construct a quadrilateral PQRS in which PQ = 3 cm, QR = 2·5 cm, PS = 3·5 cm, PR = 4 cm and QS = 5 cm.
Answer:
Steps of construction :
(i) Draw PQ = 3 cm.
(ii) With P as centre and radius = 4 cm,
draw an arc with Q as centre and radius = 2·5 cm
draw an arc to meet the previous arc at R.
Join PR and QR.
(iii) With P as centre and radius = 3·5 cm,
draw an arc. With Q as centre and radius = 5 cm,
draw an arc to meet the previous arc at S.
(iv) Join PS, QS and SR.
(v) Hence, PQRS is the required quadrilateral.
Construction of Quadrilaterals Exe-14.1
ML Aggarwal Class 8 ICSE Maths Solutions
Page-255
Question 4. Construct a quadrilateral ABCD such that BC = 5 cm, AD = 5.5 cm, CD = 4.5 cm, AC = 7 cm, and BC = 5.5 cm.
Answer:
Steps of construction :
(i) Draw a line segment CD = 4.5 cm.
(ii) With centre C and radius of 5.5 cm and with centre D
and radius 7 cm draw arcs intersecting each other at B.
(iii) Join BC and BD.
(iv) Similarly with centre C and radius 5.5 cm
and with centre D and radius 5.5 cm,
draw arcs intersecting each other at A.
(v) Join AC and AD.
(vi) Join AB.
ABCD is the required quadrilateral.
(ML Aggarwal Construction of Quadrilaterals Exe-14.1 Class 8)
Question 5. Construct a quadrilateral ABCD given that BC = 6 cm, CD = 4 cm, ∠B = 45°, ∠C = 90° and ∠D = 120°.
Answer:
Steps of construction :
(i) Draw BC = 6 cm.
(ii) At B, construct ∠CBP = 45°.
(iii) At C, construct ∠BCQ = 90°
(iv) From CQ, cut off CD = 4 cm.
(v) At D, construct ∠CDR = 120°.
(vi) Let BP and DR meet at A.
Then ABCD is the required quadrilateral.
Question 6. Construct a quadrilateral PQRS where PQ = 4 cm, QR = 6 cm, ∠P = 60°, ∠Q = 90° and ∠R = 120°.
Answer:
Steps of construction :
(i) Draw a line segment QR = 6 cm.
(ii) At Q, draw a ray QX making an angle of 90°
and cut off QP = 4 cm.
(iii) At P, draw a ray making an angle of 60° and at R,
a ray making an angle 120° which meet each other at S.
PQRS is the required quadrilateral.
Question 7. Construct a quadrilateral ABCD such that AB = 5 cm, BC = 4·2 cm, AD = 3·5 cm, ∠A = 90°, and ∠B = 60°
Answer:
Steps of construction :
(i) Draw AB = 5 cm.
(ii) At A, construct angle A = 90°
(iii) At B, construct angle B = 60°
(iv) With B as centre and 4·2 cm as radius, cut off ∠B atC.
(v) With A as centre and 3·5 cm as radius, cut off ∠A at D.
(vi) Join CD. Then ABCD is the required quadrilateral.
Question 8. Cosntruct a quadrilateral PQRS where PQ = 4 cm, QR = 5 cm, RS = 4.5 cm, ∠Q = 60° and ∠R = 90°.
Answer:
Steps of construction :
(i) Draw a line segment QR = 5 cm.
(ii) At Q, draw a ray QX making an angle of 60°
and cut off QP = 4 cm.
(iii) At R, draw a ray RY making an angle of 90°
and cut off RS. = 4.5 cm.
(iv) Join PS.
PQRS is the required quadrilateral.
(ML Aggarwal Construction of Quadrilaterals Exe-14.1 Class 8)
Question 9. Construct a quadrilateral BEST where BE = 3.8 cm, ES = 3.4 cm, ST = 4.5 cm, TB = 5 cm and ∠E = 80°.
Answer:
Steps of construction :
(i) Draw a line segment BE = 3.8 cm.
(ii) At E, draw a ray EX making an angle of 80°
and cut off ES = 3.4 cm.
(iii) With centre B and radius 5 cm and
with centre S with radius 4.5 cm,
draw arcs intersecting each other at T.
(iv) Join TB and TS.
BEST is the required quadrilateral.
Question 10. Construct a quadrilateral ABCD where AB = 4·5 cm, BC = 4 cm, CD = 3·9 cm, AD = 3·2 cm and ∠B = 60°.
Answer:
Steps of construction :
(i) Draw AB = 4·5 cm.
(ii) At B, construct ∠ABP = 60°.
(iii) From BP, cut of ∠BC = 4 cm.
(iv) With C as centre, and 3·9 cm as radius draw an arc.
(v) With A as centre and 3·2 cm as radius,
draw an arc to meet the previous arc at D.
(vi) Join AD and CD.
— End of Construction of Quadrilaterals Exe-14.1 Class 8 ICSE Maths Solutions :–
Return to : – ML Aggarwal Maths Solutions for ICSE Class -8
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