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NCERT Solutions for Class 7 Maths
NCERT Solutions for Class 7 Maths are solved by experts in order to help students to obtain excellent marks in their annual examination. All the questions and answers that are present in the CBSE NCERT Books has been included in this page. We have provided all the Class 7 Maths NCERT Solutions with a detailed explanation i.e., we have solved all the question with step by step solutions in understandable language. So students having great knowledge over NCERT Solutions Class 7 Maths can easily make a grade in their board exams.
Chapter - 10 Practical Geometry
Ex 10.1 Class 7 Maths Question 1.
Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.
Solution:
Steps of Construction :
- Take any point P on the line AB.
- Take any point C outside AB and join CP.
- With P as centre, draw an arc cutting AB and PC at X and Y respectively.
- With centre C and the same radius as in step 3, draw an arc on the opposite side of the PC to cut the PC at Q.
- With centre Q and radius equal to XY, draw an arc cutting the arc drawn in step 4 at R.
- Join CP and produce it in both directions to obtain the required line.
Ex 10.1 Class 7 Maths Question 2.
Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.
Solution:
Steps of Construction:
- Draw a line l and take any point P on it.
- With P as centre and any radius, draw an arc to intersect line l at A and B.
- With A as centre and radius greater than PA, draw in an arc.
- With centre B and the same radius, as in step 2, draw another arc to intersect the arc drawn in step 2 at C.
- Join PC and produce it to Q. Then PQ ⊥ l
- With P as centre and radius equal to 4 cm, draw an arc to intersect PQ at X such than PX = 4 cm.
- At X, make ∠RXP = ∠BPX.
- Join XR to obtain the required line m.
Validity: Since ∠BXP = ∠BPX and these are alternate angles, therefore, .XR || l, i.e., m || l and contain X such that PX = 4 cm and ∠XPB = 90°.
Ex 10.1 Class 7 Maths Question 3.
Let l be a line and P be a point on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose?
Solution:
Steps of Construction :
- Draw a line l and take any point P outside it.
- Take any point Q on line l.
- Join PQ.
- With a centre, draw an arc cutting l and PQ at C and D respectively.
- With centre P and the same radius as in step 4, draw an arc on the opposite side of PQ to cut PQ at E.
- With centre E and radius equal to CD,’ draw an arc cutting the arc of step 5 at F.
- Join PF and produce it in both directions to obtain the required line m.
- Take any point R on m.
- Through P, draw a line PS || PQ by following the steps already explained.
- The shape of the figure endorsed by these lines is a parallelogram RPQS.
Ex 10.2 Class 7 Maths Question 1.
Construct ∆XYZ in which XY = 45 cm, YZ = 5 cm and ZX = 6 cm.
Solution:
Steps of Construction :
- Draw a line segment YZ = 5 cm.
- With Y centre and draw an arc radius cm.
- With Z as centre and draw another arc intersecting the arc radius = 6 cm, at X.
- Join XY and XZ to obtain the required triangle.
Ex 10.2 Class 7 Maths Question 2.
Construct an equilateral triangle of side 5.5 cm.
Solution:
Steps of Construction :
- Draw a line segment BC = 5.5 cm.
- With centre B and radius = 5.5 cm, draw an arc.
- With centre C and radius = 5.5 cm, draw another arc intersecting the arc drawn in step 2 at A.
- Join AB and AC to obtain the required triangle.
Ex 10.2 Class 7 Maths Question 3.
Draw ∆PQR with PQ = 4, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
Solution:
Steps of Construction :
- Draw a line segment QR = 3.5 cm.
- With centre Q and radius = 4 cm, draw an arc.
- With R as centre and radius = 4 cm, draw another arc intersecting the arc drawn in step 2 at P.
- Join PQ and PR to obtain the required triangle. ∆PQR is an isosceles triangle.
Ex 10.2 Class 7 Maths Question 4.
Construct AABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Solution:
Steps of Construction :
- Draw a line segment BC = 6 c.m
- With centre B and radius = 2.5 cm, draw an arc.
- With centre C and radius =6.5 cm, draw another arc intersecting the arc drawn in step 2 at A.
- Join AB and AC to obtain the required triangle.
On measuring, we find that ∠B = 90°.
Ex 10.3 Class 7 Maths Question 1.
Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m ∠EDF = 90°
Solution:
Steps of Constipation :
- Draw a line segmerit DE = 5cm.
- Draw ∠EDX = 90°.
- With centre D and radius = 3 cm, draw an are to intersect DX at F.
- Join EF to obtain the required triangle DBF.
Ex 10.3 Class 7 Maths Question 2.
Construct an isosceles triangle in which the lengths of each of its equal Sides is 6.5 cm find the angle between them is 110°
Solution:
Steps of Construction :
- Draw a line segment BC = 6.5 cm.
- Draw ∠CBX = 110°.
- With B as centre and radius = 6.5 cm, draw an arc intersecting BX at A.
- Join AC to obtain the required ∆ABC.
Ex 10.3 Class 7 Maths Question 3.
Construct ∆ABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.
Solution:
Steps of Construction :
- Draw a line segment BC =7.5 cm.
- Draw ∠BCX = 60°.
- With C as centre and radius = 5 cm, draw an arc intersecting CX at A.
- Join AB to obtain the required ∆ABC.
Ex 10.4 Class 7 Maths Question 1.
Construct ∆ABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.
Solution:
Steps of Construction :
1. Draw a line segment AB = 5.8 cm.
2. Draw ∠BAX = 60°.
3. Draw ∠ABY, with Y on the same side of AB
such that ∠ABY = 30°.
Let AX and BY interest at C.
Then, ∆ABC is the required triangle.
Ex 10.4 Class 7 Maths Question 2.
Construct ∆PQR if PQ = 5cm, m ∠PQR = 105° and m ∠QRP = 40°.
Solution:
Here, we are given the side PQ, ∠Q and ∠R. But to draw the triangle, we require ∠P
Ex 10.4 Class 7 Maths Question 3.
Examine whether you can construct
∆DEF such that EF = 7.2 cm, m∠E = 110° and m∠F = 80°. Justify your answer.
Solution:
Since m∠E +m∠F = 110° + 80° = 190°, so the ∆DEF cannot be drawn as the’sum of all the angles of a triangle is 180°.
Ex 10.5 Class 7 Maths Question 1.
Construct the. right-angled ∆PQR, where m∠Q = 90°, QR = 8 cm and PR = 10 cm.
Solution:
Steps of Construction :
- Draw a line segment QR = 8 cm.
- Draw ∠XQR = 90°.
- With R as centre and radius =10 cm, draw an arc to intersect ray QX at P.
- Join RP to obtain the required ∆PQR.
Ex 10.5 Class 7 Maths Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Solution:
Steps of Construction :
- Draw a line segment QR = 4 cm.
- Draw ∠XQR = 90°.
- With R as centre and radius equal to hypotenuse 6 cm, draw an arc to intersect ray QX at P.
- Join RP to obtain the required ∆PQR.
Ex 10.5 Class 7 Maths Question 3.
Construct an isosceles right-angled triangle ABC, where m ∠ACB = 90° and AC = 6 cm.
Solution:
Steps of Construction :
- Draw a line segment CB = 6 cm (∵ CB = AC = 6 cm)
- Draw ∠BCX = 90°.
- With C as centre and radius = 6 cm, draw an arc to intersect ray CX at A.
- Join BA to obtain the required triangle ABC.
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