Given radius r = 7 cm and sector angle θ = 60°. The formula for the area of a sector is: Area = (θ / 360) x π x r² Substitute the given values using π = 22/7: Area = (60 / 360) x (22/7) x 7 x 7 Area = (1/6) x 22 x 7 Area = 154 / 6 = 77 / 3 cm². In decimal form, 77 / 3 is approximately 25.67 cm². TheRead more
Given radius r = 7 cm and sector angle θ = 60°.
The formula for the area of a sector is:
Area = (θ / 360) x π x r²
Substitute the given values using π = 22/7:
Area = (60 / 360) x (22/7) x 7 x 7
Area = (1/6) x 22 x 7
Area = 154 / 6 = 77 / 3 cm².
In decimal form, 77 / 3 is approximately 25.67 cm².
Therefore, the area of the sector is 77/3 cm² (or 25.67 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Given CQ is parallel to PD. Triangles DPQ and DPC lie on the same base PD and between the same parallel lines PD and CQ. Therefore, their areas are equal: Area(ΔDPQ) = Area(ΔDPC). Now, consider triangle BPQ: Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ) Substitute Area(ΔDPC) for Area(ΔDPQ): Area(ΔBPQ) = AreaRead more
Given CQ is parallel to PD.
Triangles DPQ and DPC lie on the same base PD and between the same parallel lines PD and CQ.
Let the side length of square ABCD be s. Draw a line through point P perpendicular to AB and CD. Let the perpendicular distance from P to AB be h1 and to CD be h2. Then h1 + h2 = s. Area of red region = Area(ΔPAB) + Area(ΔPCD) = (1/2) x s x h1 + (1/2) x s x h2 = (1/2) x s x (h1 + h2) = (1/2) x s². SRead more
Let the side length of square ABCD be s.
Draw a line through point P perpendicular to AB and CD.
Let the perpendicular distance from P to AB be h1 and to CD be h2.
Then h1 + h2 = s.
Area of red region = Area(ΔPAB) + Area(ΔPCD)
= (1/2) x s x h1 + (1/2) x s x h2 = (1/2) x s x (h1 + h2) = (1/2) x s².
Since total area of the square is s², the green region also equals s² – s²/2 = (1/2) x s².
Thus, the ratio of areas is 1:1.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Since D is the midpoint of BC, AD is a median of triangle ABC. A median divides a triangle into two triangles of equal area: Area(ΔABD) = Area(ΔACD) ... (Equation 1). In triangle PBC, PD is also a median because D is the midpoint of BC. Therefore: Area(ΔPBD) = Area(ΔPCD) ... (Equation 2). SubtractinRead more
Since D is the midpoint of BC, AD is a median of triangle ABC.
A median divides a triangle into two triangles of equal area:
Area(ΔABD) = Area(ΔACD) … (Equation 1).
In triangle PBC, PD is also a median because D is the midpoint of BC.
Therefore:
Area(ΔPBD) = Area(ΔPCD) … (Equation 2).
Subtracting Equation 2 from Equation 1:
Area(ΔABD) – Area(ΔPBD) = Area(ΔACD) – Area(ΔPCD)
Area(ΔABP) = Area(ΔACP).
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Let ABCD be the 4-gon and E, F, G, H the midpoints of AB, BC, CD, DA respectively. Draw diagonal AC, dividing ABCD into triangles ABC and ADC. In triangle ABC, E and F are midpoints, so triangle EBF is similar to ABC with ratio 1/2, meaning Area(EBF) = (1/4) x Area(ABC). Similarly, Area(HDG) = (1/4)Read more
Let ABCD be the 4-gon and E, F, G, H the midpoints of AB, BC, CD, DA respectively.
Draw diagonal AC, dividing ABCD into triangles ABC and ADC.
In triangle ABC, E and F are midpoints, so triangle EBF is similar to ABC with ratio 1/2, meaning Area(EBF) = (1/4) x Area(ABC).
Similarly, Area(HDG) = (1/4) x Area(ADC).
Adding these:
Area(EBF) + Area(HDG) = (1/4) x [Area(ABC) + Area(ADC)] = (1/4) x Area(ABCD).
Similarly, using diagonal BD:
Area(HAE) + Area(GCF) = (1/4) x Area(ABCD).
Sum of four corner triangles = (1/4 + 1/4) x Area(ABCD) = (1/2) x Area(ABCD).
Subtracting corners gives:
Area(EFGH) = Area(ABCD) – (1/2) x Area(ABCD) = (1/2) x Area(ABCD).
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Given radius r = 7 cm and sector angle θ = 60°. The formula for the area of a sector is: Area = (θ / 360) x π x r² Substitute the given values using π = 22/7: Area = (60 / 360) x (22/7) x 7 x 7 Area = (1/6) x 22 x 7 Area = 154 / 6 = 77 / 3 cm². In decimal form, 77 / 3 is approximately 25.67 cm². TheRead more
Given radius r = 7 cm and sector angle θ = 60°.
The formula for the area of a sector is:
Area = (θ / 360) x π x r²
Substitute the given values using π = 22/7:
Area = (60 / 360) x (22/7) x 7 x 7
Area = (1/6) x 22 x 7
Area = 154 / 6 = 77 / 3 cm².
In decimal form, 77 / 3 is approximately 25.67 cm².
Therefore, the area of the sector is 77/3 cm² (or 25.67 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessIn ΔABC, D is the midpoint of AB. P is any point on BC and Q is a point on AB such that CQ || PD. PQ is joined (Fig. 6.34). Prove that Area (ΔBPQ) = 1/2 Area (ΔABC).
Given CQ is parallel to PD. Triangles DPQ and DPC lie on the same base PD and between the same parallel lines PD and CQ. Therefore, their areas are equal: Area(ΔDPQ) = Area(ΔDPC). Now, consider triangle BPQ: Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ) Substitute Area(ΔDPC) for Area(ΔDPQ): Area(ΔBPQ) = AreaRead more
Given CQ is parallel to PD.
Triangles DPQ and DPC lie on the same base PD and between the same parallel lines PD and CQ.
Therefore, their areas are equal:
Area(ΔDPQ) = Area(ΔDPC).
Now, consider triangle BPQ:
Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ)
Substitute Area(ΔDPC) for Area(ΔDPQ):
Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPC) = Area(ΔBDC).
Since D is the midpoint of AB, CD is a median of triangle ABC.
A median divides the triangle into two equal halves:
Area(ΔBDC) = (1/2) x Area(ΔABC).
Hence, Area(ΔBPQ) = (1/2) x Area(ΔABC).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessGiven a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (ΔPAB and ΔPCD) and the green region (ΔPBC and ΔPDA)?
Let the side length of square ABCD be s. Draw a line through point P perpendicular to AB and CD. Let the perpendicular distance from P to AB be h1 and to CD be h2. Then h1 + h2 = s. Area of red region = Area(ΔPAB) + Area(ΔPCD) = (1/2) x s x h1 + (1/2) x s x h2 = (1/2) x s x (h1 + h2) = (1/2) x s². SRead more
Let the side length of square ABCD be s.
Draw a line through point P perpendicular to AB and CD.
Let the perpendicular distance from P to AB be h1 and to CD be h2.
Then h1 + h2 = s.
Area of red region = Area(ΔPAB) + Area(ΔPCD)
= (1/2) x s x h1 + (1/2) x s x h2 = (1/2) x s x (h1 + h2) = (1/2) x s².
Since total area of the square is s², the green region also equals s² – s²/2 = (1/2) x s².
Thus, the ratio of areas is 1:1.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessIn ΔABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area (ΔABP) = area (ΔACP).
Since D is the midpoint of BC, AD is a median of triangle ABC. A median divides a triangle into two triangles of equal area: Area(ΔABD) = Area(ΔACD) ... (Equation 1). In triangle PBC, PD is also a median because D is the midpoint of BC. Therefore: Area(ΔPBD) = Area(ΔPCD) ... (Equation 2). SubtractinRead more
Since D is the midpoint of BC, AD is a median of triangle ABC.
A median divides a triangle into two triangles of equal area:
Area(ΔABD) = Area(ΔACD) … (Equation 1).
In triangle PBC, PD is also a median because D is the midpoint of BC.
Therefore:
Area(ΔPBD) = Area(ΔPCD) … (Equation 2).
Subtracting Equation 2 from Equation 1:
Area(ΔABD) – Area(ΔPBD) = Area(ΔACD) – Area(ΔPCD)
Area(ΔABP) = Area(ΔACP).
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessIf the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)
Let ABCD be the 4-gon and E, F, G, H the midpoints of AB, BC, CD, DA respectively. Draw diagonal AC, dividing ABCD into triangles ABC and ADC. In triangle ABC, E and F are midpoints, so triangle EBF is similar to ABC with ratio 1/2, meaning Area(EBF) = (1/4) x Area(ABC). Similarly, Area(HDG) = (1/4)Read more
Let ABCD be the 4-gon and E, F, G, H the midpoints of AB, BC, CD, DA respectively.
Draw diagonal AC, dividing ABCD into triangles ABC and ADC.
In triangle ABC, E and F are midpoints, so triangle EBF is similar to ABC with ratio 1/2, meaning Area(EBF) = (1/4) x Area(ABC).
Similarly, Area(HDG) = (1/4) x Area(ADC).
Adding these:
Area(EBF) + Area(HDG) = (1/4) x [Area(ABC) + Area(ADC)] = (1/4) x Area(ABCD).
Similarly, using diagonal BD:
Area(HAE) + Area(GCF) = (1/4) x Area(ABCD).
Sum of four corner triangles = (1/4 + 1/4) x Area(ABCD) = (1/2) x Area(ABCD).
Subtracting corners gives:
Area(EFGH) = Area(ABCD) – (1/2) x Area(ABCD) = (1/2) x Area(ABCD).
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See less