In Fig. 6.43, let the vertices of the main triangle be A (top), B (bottom-left), C (bottom-right). The left side is bisected (ratio 1:1) and the right side is trisected (ratio 1:1:1). The top unshaded triangle has area (1/2) x (1/3) = 1/6 of the total area. T he bottom unshaded triangle has base aloRead more
In Fig. 6.43, let the vertices of the main triangle be A (top), B (bottom-left), C (bottom-right).
The left side is bisected (ratio 1:1) and the right side is trisected (ratio 1:1:1).
The top unshaded triangle has area (1/2) x (1/3) = 1/6 of the total area. T
he bottom unshaded triangle has base along the bottom and vertex at the 2/3 mark along AC, covering (1/3) of the total area.
Shaded fraction = 1 – 1/6 – 1/3 = 1 – 3/6 = 1/2 (or 7/18 depending on line endpoints; standard dissection gives 7/18).
In Fig. 6.44, lines join each vertex of the square to the midpoint of an opposite side. By translating the four surrounding right-angled triangles into the inner shape, the shaded square has area exactly 1/5 of the total square.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
(i) Area(ABCD) = ab. Area(PQRS) = (2a) x (2b) = 4ab = 4 x Area(ABCD). By dividing each side of PQRS into two equal parts, it divides into a 2-by-2 grid of 4 identical copies of ABCD that fit perfectly. (ii) By Heron's formula or similarity, scaling sides by 2 scales area by 2² = 4. Joining the midpoRead more
(i) Area(ABCD) = ab. Area(PQRS) = (2a) x (2b) = 4ab = 4 x Area(ABCD).
By dividing each side of PQRS into two equal parts, it divides into a 2-by-2 grid of 4 identical copies of ABCD that fit perfectly.
(ii) By Heron’s formula or similarity, scaling sides by 2 scales area by 2² = 4.
Joining the midpoints of the sides of ΔPQR creates the medial triangle and divides ΔPQR into 4 congruent copies of ΔABC, which fit seamlessly.
(iii) Scaling sides by 3 scales area by 3² = 9.
Trisecting each side of ΔPQR and drawing grid lines parallel to the sides subdivides ΔPQR into 1 + 3 + 5 = 9 identical copies of ΔABC that fit completely without overlap.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
In a kite, diagonals d1 and d2 are perpendicular and one diagonal (say d1) bisects the other (d2 = h1 + h2). (i) Using algebra: The main diagonal d1 divides the kite into two triangles with common base d1 and perpendicular heights h1 and h2. Total Area = (1/2)d1h1 + (1/2)d1h2 = (1/2)d1(h1 + h2) = (1Read more
In a kite, diagonals d1 and d2 are perpendicular and one diagonal (say d1) bisects the other (d2 = h1 + h2).
(i) Using algebra:
The main diagonal d1 divides the kite into two triangles with common base d1 and perpendicular heights h1 and h2.
Total Area = (1/2)d1h1 + (1/2)d1h2 = (1/2)d1(h1 + h2) = (1/2)d1d2.
(ii) Using geometry:
Draw lines through the vertices parallel to both diagonals to form a bounding rectangle of dimensions d1 and d2 (area = d1d2). Each of the four corner right triangles outside the kite is congruent to an adjacent right triangle inside the kite. Thus, the kite fills exactly half the rectangle:
Area = (1/2)d1d2.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Take two congruent trapeziums, each with parallel sides a and b and height h. Rotate the second trapezium by 180 degrees and place its non-parallel side against the corresponding side of the first trapezium. The top side becomes a + b and the bottom side becomes b + a = a + b, with the same perpendiRead more
Take two congruent trapeziums, each with parallel sides a and b and height h.
Rotate the second trapezium by 180 degrees and place its non-parallel side against the corresponding side of the first trapezium.
The top side becomes a + b and the bottom side becomes b + a = a + b, with the same perpendicular distance h.
This combined figure forms a parallelogram with base (a + b) and height h.
Area of the parallelogram = base x height = (a + b)h.
Since it contains two identical trapeziums:
Area of one trapezium = (1/2) x Area of parallelogram = (1/2)(a + b)h.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Let the parallel sides of the trapezium have lengths a and b and let the perpendicular distance between them be h. Drawing a diagonal connects opposite vertices and divides the trapezium into two non-overlapping triangles: The first triangle has base a and altitude h, giving area equal to (1/2) x aRead more
Let the parallel sides of the trapezium have lengths a and b and let the perpendicular distance between them be h.
Drawing a diagonal connects opposite vertices and divides the trapezium into two non-overlapping triangles:
The first triangle has base a and altitude h, giving area equal to (1/2) x a x h.
The second triangle has base b and altitude h, giving area equal to (1/2) x b x h.
Total area of trapezium = Area of first triangle + Area of second triangle
= (1/2)ah + (1/2)bh = (1/2)(a + b)h.
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Fig. 6.43: What fraction of the triangle is shaded? Fig. 6.44: What fraction of the square is shaded?
In Fig. 6.43, let the vertices of the main triangle be A (top), B (bottom-left), C (bottom-right). The left side is bisected (ratio 1:1) and the right side is trisected (ratio 1:1:1). The top unshaded triangle has area (1/2) x (1/3) = 1/6 of the total area. T he bottom unshaded triangle has base aloRead more
In Fig. 6.43, let the vertices of the main triangle be A (top), B (bottom-left), C (bottom-right).
The left side is bisected (ratio 1:1) and the right side is trisected (ratio 1:1:1).
The top unshaded triangle has area (1/2) x (1/3) = 1/6 of the total area. T
he bottom unshaded triangle has base along the bottom and vertex at the 2/3 mark along AC, covering (1/3) of the total area.
Shaded fraction = 1 – 1/6 – 1/3 = 1 – 3/6 = 1/2 (or 7/18 depending on line endpoints; standard dissection gives 7/18).
In Fig. 6.44, lines join each vertex of the square to the midpoint of an opposite side. By translating the four surrounding right-angled triangles into the inner shape, the shaded square has area exactly 1/5 of the total square.
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 lessThree problems about fitting congruent shapes together: (i) Rectangle ABCD has sides a, b and rectangle PQRS has sides 2a, 2b. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see! (ii) ΔABC has sides a, b, c and ΔPQR has sides 2a, 2b, 2c. Show that ΔPQR has 4 times the area of ΔABC. Does this mean that 4 copies of ΔABC will fit into ΔPQR? Check and see! (iii) ΔABC has sides a, b, c and ΔPQR has sides 3a, 3b, 3c. Show that ΔPQR has 9 times the area of ΔABC. Does this mean that 9 copies of ΔABC will fit into ΔPQR? Check and see!
(i) Area(ABCD) = ab. Area(PQRS) = (2a) x (2b) = 4ab = 4 x Area(ABCD). By dividing each side of PQRS into two equal parts, it divides into a 2-by-2 grid of 4 identical copies of ABCD that fit perfectly. (ii) By Heron's formula or similarity, scaling sides by 2 scales area by 2² = 4. Joining the midpoRead more
(i) Area(ABCD) = ab. Area(PQRS) = (2a) x (2b) = 4ab = 4 x Area(ABCD).
By dividing each side of PQRS into two equal parts, it divides into a 2-by-2 grid of 4 identical copies of ABCD that fit perfectly.
(ii) By Heron’s formula or similarity, scaling sides by 2 scales area by 2² = 4.
Joining the midpoints of the sides of ΔPQR creates the medial triangle and divides ΔPQR into 4 congruent copies of ΔABC, which fit seamlessly.
(iii) Scaling sides by 3 scales area by 3² = 9.
Trisecting each side of ΔPQR and drawing grid lines parallel to the sides subdivides ΔPQR into 1 + 3 + 5 = 9 identical copies of ΔABC that fit completely without overlap.
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 lessShow that the area of a kite is half the product of its diagonals. Show this: (i) using algebra and (ii) using geometry.
In a kite, diagonals d1 and d2 are perpendicular and one diagonal (say d1) bisects the other (d2 = h1 + h2). (i) Using algebra: The main diagonal d1 divides the kite into two triangles with common base d1 and perpendicular heights h1 and h2. Total Area = (1/2)d1h1 + (1/2)d1h2 = (1/2)d1(h1 + h2) = (1Read more
In a kite, diagonals d1 and d2 are perpendicular and one diagonal (say d1) bisects the other (d2 = h1 + h2).
(i) Using algebra:
The main diagonal d1 divides the kite into two triangles with common base d1 and perpendicular heights h1 and h2.
Total Area = (1/2)d1h1 + (1/2)d1h2 = (1/2)d1(h1 + h2) = (1/2)d1d2.
(ii) Using geometry:
Draw lines through the vertices parallel to both diagonals to form a bounding rectangle of dimensions d1 and d2 (area = d1d2). Each of the four corner right triangles outside the kite is congruent to an adjacent right triangle inside the kite. Thus, the kite fills exactly half the rectangle:
Area = (1/2)d1d2.
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 lessShow how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?
Take two congruent trapeziums, each with parallel sides a and b and height h. Rotate the second trapezium by 180 degrees and place its non-parallel side against the corresponding side of the first trapezium. The top side becomes a + b and the bottom side becomes b + a = a + b, with the same perpendiRead more
Take two congruent trapeziums, each with parallel sides a and b and height h.
Rotate the second trapezium by 180 degrees and place its non-parallel side against the corresponding side of the first trapezium.
The top side becomes a + b and the bottom side becomes b + a = a + b, with the same perpendicular distance h.
This combined figure forms a parallelogram with base (a + b) and height h.
Area of the parallelogram = base x height = (a + b)h.
Since it contains two identical trapeziums:
Area of one trapezium = (1/2) x Area of parallelogram = (1/2)(a + b)h.
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 lessBy dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).
Let the parallel sides of the trapezium have lengths a and b and let the perpendicular distance between them be h. Drawing a diagonal connects opposite vertices and divides the trapezium into two non-overlapping triangles: The first triangle has base a and altitude h, giving area equal to (1/2) x aRead more
Let the parallel sides of the trapezium have lengths a and b and let the perpendicular distance between them be h.
Drawing a diagonal connects opposite vertices and divides the trapezium into two non-overlapping triangles:
Total area of trapezium = Area of first triangle + Area of second triangle
= (1/2)ah + (1/2)bh = (1/2)(a + b)h.
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