A PQR in which PQ = 5.5 cm, QR = 6.5 cm, ZQ = 40°.
Answers
Mathematics_(solutions) Solutions for Class 7 Math Chapter 1 Geometrical Constructions are provided here with simple step-by-step explanations. These solutions for Geometrical Constructions are extremely popular among Class 7 students for Math Geometrical Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics_(solutions) Book of Class 7 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics_(solutions) Solutions. All Mathematics_(solutions) Solutions for class Class 7 Math are prepared by experts and are 100% accurate.
Page No 2:
Question 1:
Draw line segments of the lengths given below and draw their perpendicular bisectors.
(3) 3.8 cm
Steps of construction:
1. Draw line segment XY = 3.8 cm.
2. With X as centre and radius more than half of XY, mark two arcs, one above and other below the line XY.
3. With Y as centre and same radius, draw two arcs cutting the previous drawn arcs and name the point of intersection as A and B.
4. Join AB and name the point where this line cuts XY as point R.
AB is the perpendicular bisector of XY.
Page No 2:
Question 2:
Draw angles of the measures given below and draw their bisectors.
(1) 105° (2) 55° (3) 90°
ANSWER:
(1) 105°
Steps of Construction:
1. Draw a ray BC.
2. With B as centre, use a protractor to make an angle of 105°. Thus, ∠ABC = 105°.
3. With X and Y as centre, draw arcs intersecting each other at point M.
BM is the required angle bisector of ∠ABC.
(2) 55°
Steps of Construction:
1. Draw a ray OX.
2. With O as centre, use a protractor to make an angle of 55°. Thus, ∠YOX = 55°.
3. With P and Q as centre, draw arcs intersecting each other at point A.
OA is the required angle bisector of ∠YOX.
(3) 90°
Steps of Construction:
1. Draw a ray OB.
2. With O as centre, use a protractor to make an angle of 90°. Thus, ∠AOB = 90°.
3. With X and Y as centre, draw arcs intersecting each other at point S.
OS is the required angle bisector of ∠AOB.
Page No 2:
Question 3:
Draw an obtuse-angled triangle and a right-angled triangle. Find the points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie?
ANSWER:
I. Right angled triangle
Steps of construction
1. Draw a right angled triangle ABC, right angled at B.
2. Make the angle bisectors of the angles A, B and C.
The angle bisectors meet at the point O. This point of concurrence of the angle bisectors lies inside the triangle ABC.
II. Obtuse-angled triangle
Steps of construction
1. Draw an obtuse angled triangle XYZ.
2. Make the angle bisectors of angles X, Y and Z.
The angle bisectors meet at point S. This point of concurrence of the angle bisectors lies inside the obtuse angled triangle XYZ.
Page No 2:
Question 4:
Draw a right-angled triangle. Draw the perpendicular bisectors of its sides. Where does the point of concurrence lie ?
ANSWER:
Steps of construction
1. Draw the right angle triangle ABC.
2. Draw the perpendicular bisectors of the sides AB, BC and CA.
The perpendicular bisectors meet at the point D which lies on the hypotenuse AC.
Page No 2:
Question 5:
Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.
ANSWER:
Maithili, Shaila and Ajay be the three vertices of a triangle.
A toy shop equidistant from these three points will be the point of concurrence of the perpendicular bisectors
of the lines joining the three vertices of the triangle. Thus, the geometrical construction representing this will be the
circumcircle.
Page No 4:
Question 1:
Draw triangles with the measures given
below.
(a) In ΔABC , l(AB) = 5.5 cm,l(BC) = 4.2 cm, l(AC) = 3.5 cm
(b) In ∆ STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm
(c) In ∆ PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm
ANSWER:
(a) In ΔABC , l(AB) = 5.5 cm,l(BC) = 4.2 cm, l(AC) = 3.5 cm
Steps of construction
1. Draw a line AB = 5.5 cm
2. With A as centre and 3.5 cm as the radius, draw an arc above the line AB.
3. With B as the centre and 4.2 cm as the radius, draw an arc cutting the previous drawn arc at point C.
4. Join CA and CB.
ΔABC is thus the required triangle.
(b) In ∆ STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm
Steps of construction
1. Draw a line ST = 7 cm
2. With S as centre and 5 cm as the radius, draw an arc above the line ST.
3. With T as the centre and 4 cm as the radius, draw an arc cutting the previous drawn arc at point U.
4. Join US and UT.
ΔSTU is thus the required triangle.
(c) In ∆ PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm
Steps of construction
1. Draw a line PQ = 6 cm
2. With P as centre and 4.5 cm as the radius, draw an arc above the line PQ.
3. With Q as the centre and 3.8 cm as the radius, draw an arc cutting the previous drawn arc at point R.
4. Join RP and RQ.
ΔPQR is thus the required triangle.