Class 9 Maths Ganita Manjari Chapter 6 Solutions — Measuring Space: Perimeter and Area

Only boundary arcs and exposed straight edges count; dotted construction lines do not. The equal-length tick marks in Fig. 6.14 determine the radii.

1. Two straights of 80 m and two semicircles of diameter 60 m: $160+60\pi \approx 348.57m$.
2. Semicircular arcs with radii 6 and 4 cm, plus two straight widths of 2 cm: $10\pi +4\approx 35.43\,\mathrm{cm}$.
3. Four semicircles of diameter 10 cm: $20\pi \approx 62.86\,\mathrm{cm}$.
4. Three semicircles of diameter 12 cm: $18\pi \approx 56.57\,\mathrm{cm}$.
5. Four quarter-circle arcs of radius 14 cm and four semicircles of diameter 14 cm: $4\left(\pi \cdot \frac{14}{2}\right)+4\left(\pi \cdot 7\right)=56\pi \approx 176\,\mathrm{cm}$.
6. Large semicircle of diameter 28 cm plus four smaller semicircles whose diameters add to 28 cm: $28\pi \approx 88\,\mathrm{cm}$.
7. The hypotenuse is $\sqrt{8^{2}+6^{2}}=10\,\mathrm{cm}$. Three semicircle diameters are 8,6,10 cm: $12\pi \approx 37.71\,\mathrm{cm}$.
8. Large diameter 12 cm; three small diameters of 4 cm: $12\pi \approx 37.71\,\mathrm{cm}$.
9. Large diameter 20 cm and two small diameters of 10 cm: $20\pi \approx 62.86\,\mathrm{cm}$.

Assuming rolling without slipping: (i) distance per turn =$\pi d\approx \left(\frac{22}{7}\right)\times 56=176\,\mathrm{cm}$. (ii) $10\,\mathrm{km}=1{,}000{,}000\,\mathrm{cm}$, so turns =$$\begin{aligned}\frac{1{,}000{,}000}{176}&=\frac{62500}{11}\\&\approx 5681.82\text{.}\end{aligned}$$ This includes a fractional final turn; about 5682 is the nearest whole-number estimate.

(i) The square has side 14 cm. Eight quarter-circle arcs have radius 7 cm. Their total length is $8\times \left(\frac{1}{4}\right)\times 2\pi \times 7=28\pi \approx 88\,\mathrm{cm}$.

(ii) The regular hexagon has side 42 cm. Each of six petals has two 60° arcs of radius 42 cm, centred at neighbouring vertices. Total length =$$\begin{aligned}12\times \left(\frac{60}{360}\right)\times 2\pi \times 42&=168\pi \\&\approx 528\,\mathrm{cm}\text{.}\end{aligned}$$

In Fig. 6.31, ABCD is a 10 cm by 8 cm rectangle. The 8 cm label refers to the entire BC, not EC. Take $\mathrm{AD}=8\,\mathrm{cm}$ as the triangle’s base. Since E lies on BC, its perpendicular distance from AD is 10 cm. Therefore $$\begin{aligned}\text{area}\left(\mathrm{ADE}\right)&=\left(\frac{1}{2}\right)\times 8\times 10\\&=40\,\mathrm{cm}^{2}\text{,}\end{aligned}$$ regardless of E’s position on BC.

For a convex quadrilateral ABCD, let E,F,G,H be the successive side midpoints. In triangle ABC, EF is parallel to AC and equals $\frac{\mathrm{AC}}{2}$; in ADC, HG has the same direction and length. Similarly EH and FG are parallel and equal to $\frac{\mathrm{BD}}{2}$. Hence EFGH is a parallelogram.

Triangle BEF is similar to BAC with scale $\frac{1}{2}$, so its area is $\frac{1}{4}\text{area}\left(\mathrm{BAC}\right)$. Likewise DGH has area $\frac{1}{4}\text{area}\left(\mathrm{DAC}\right)$. Together these are $\frac{1}{4}$ of ABCD. The other two corner triangles have total $\frac{1}{4}$ of ABCD, using diagonal BD. Subtract the four corners: $\text{area}\left(\mathrm{EFGH}\right)=\frac{1}{2}\text{area}\left(\mathrm{ABCD}\right)$. The area result extends to a simple concave quadrilateral by signed triangle-area addition; crossed quadrilaterals need a stated signed-area convention.

Triangles BDP and BQC are similar: BP and BC are collinear, BD and BQ are collinear, and PD∥CQ. Therefore $\frac{\mathrm{BP}}{\mathrm{BC}}=\frac{\mathrm{BD}}{\mathrm{BQ}}$, so $\mathrm{BQ}\cdot \mathrm{BP}=\mathrm{BD}\cdot \mathrm{BC}$.

The perpendicular heights from P and C to line AB are in ratio BP:BC. Thus $$\begin{aligned}\frac{\text{area}\left(\mathrm{BPQ}\right)}{\text{area}\left(\mathrm{ABC}\right)}&=\left(\frac{\mathrm{BQ}}{\mathrm{BA}}\right)\left(\frac{\mathrm{BP}}{\mathrm{BC}}\right)\\&=\frac{\mathrm{BD}}{\mathrm{BA}}\\&=\frac{1}{2}\text{.}\end{aligned}$$

Domain of the construction: $P\ne B$. Q is on the line AB, possibly on its extension. To keep Q on the segment AB as drawn, P must lie at least halfway from B to C. The printed “any point” needs this qualification.

Printed-formula note: the textbook places π outside the entire bracket; that expression would be negative. The correct expression is $r^{2}\left(\frac{\pi }{6}-\frac{\sqrt{3}}{4}\right)$. Sector area =$\frac{\pi r^{2}}{6}$. The triangle is equilateral with side r and area $\frac{\sqrt{3}r^{2}}{4}$. Subtracting gives the stated corrected result.

The intended hexagon must be regular; an arbitrary inscribed hexagon does not have this area. Joining its centre to all six vertices gives six equilateral triangles of side r. Its area is $6\left(\frac{\sqrt{3}r^{2}}{4}\right)=\frac{3\sqrt{3}r^{2}}{2}$, so its fraction of the circle is $\frac{3\sqrt{3}}{2\pi }\approx 0.827$. The inscribed equilateral triangle has area $\frac{3\sqrt{3}r^{2}}{4}$, exactly half.

For geometric lengths assume $a>b>0$ in the difference-of-squares model. The algebraic identities themselves hold for all real values.

Assuming no slipping, distance per turn =$2\pi \times 28\approx 176\,\mathrm{cm}$. For $1\,\mathrm{km}=100{,}000\,\mathrm{cm}$, number of turns =$\frac{100000}{176}=\frac{6250}{11}\approx 568.18$ turns. This includes the partial final turn; 568 is only the nearest integer.

The line drawn in Fig. 6.42 splits the trapezium into a parallelogram of base a and height h, and a triangle of base b−a and the same height h. Their areas add to $ah+\left(\frac{1}{2}\right)\left(b-a\right)h=\frac{\left(a+b\right)h}{2}$.

(i) Let perpendicular diagonals have lengths $d_{1}$,$d_{2}$. Splitting along $d_{1}$ gives two triangles with perpendicular heights p and q, where $p+q=d_{2}$. Their combined area is $$\left(\frac{1}{2}\right)d_{1}\left(p+q\right)=\frac{d_{1}d_{2}}{2}\text{.}$$

(ii) Draw a rectangle through the four vertices with sides parallel to the diagonals. The diagonals partition that rectangle into four smaller rectangles. Each of the four pieces of the kite is a right triangle occupying exactly half one small rectangle. Therefore the kite fills half the surrounding $d_{1}\times d_{2}$ rectangle.

(i) Area ratio =$\frac{2a\cdot 2b}{ab}=4$. Divide both long rectangle sides at their midpoints to make a $2\times 2$ arrangement of four copies.

(ii) The side scale factor is 2, so the height also doubles and area becomes four times as large. Join the three side midpoints of the larger triangle: the four resulting triangles are congruent to the original by SSS.

(iii) Trisect every side of the larger triangle and draw the parallels to its sides through the division points. This triangular grid makes nine congruent triangles, each with side lengths a,b,c. Thus the ratio is 9 and all nine copies fit. Equal area alone would not prove a tiling; these constructions do.

(i) Label the top vertex A and the bottom vertices B,C. Let M be the midpoint of AB and N,P the trisection points of AC in that order. The unshaded top triangle AMN has fraction $\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)=\frac{1}{6}$ of ABC; the unshaded bottom triangle BPC has fraction $\frac{1}{3}$ (same altitude to AC). The shaded fraction is $1-\frac{1}{6}-\frac{1}{3}=\frac{1}{2}$.

(ii) Use a unit square with bottom-left (0,0) and top-right (1,1). The four drawn lines are $y=\frac{x}{2}$, $y=\frac{x}{2}+\frac{1}{2}$, $y=1-2x$ and $y=2-2x$. The central vertices are $\left(\frac{2}{5},\frac{1}{5}\right)$,$\left(\frac{4}{5},\frac{2}{5}\right)$,$\left(\frac{3}{5},\frac{4}{5}\right)$,$\left(\frac{1}{5},\frac{3}{5}\right)$. Adjacent side vectors are $\left(\frac{2}{5},\frac{1}{5}\right)$ and $\left(-\frac{1}{5},\frac{2}{5}\right)$, which are perpendicular and each have squared length $\frac{1}{5}$. The central square therefore occupies $\frac{1}{5}$ of the original.

For Fig. 6.45, the three radius-r circles have area $3\pi r^{2}$, while the rectangle has dimensions $6r\times 2r$, area $12r^{2}$. Fraction=$\frac{\pi }{4}$. For Fig. 6.46, the four circles have area $4\pi r^{2}$ in an $8r\times 2r$ rectangle of area $16r^{2}$, again $\frac{\pi }{4}$. Using $\frac{22}{7}$ gives $\frac{11}{14}\approx 78.57$%.

Conjecture for this one-row arrangement: n equal circles tangent to their neighbours and tightly fitted between the rectangle’s sides cover $\frac{\pi }{4}$ of it. The circles have area $n\pi r^{2}$ and the rectangle $2nr\times 2r=4nr^{2}$, so the ratio is $\frac{\pi }{4}$. For $n=10$,20,50 the respective ratios are $\frac{10\pi }{40}$,$\frac{20\pi }{80}$,$\frac{50\pi }{200}$, all $\frac{\pi }{4}$. This is not a claim about arbitrary circle packings.

Let each small rectangle have long side l and short side b. The top row has four long sides across; the bottom has five short sides across. Hence $4l=5b$, not $l=2b$. Each small area is $\frac{72}{9}=8\,\mathrm{cm}^{2}$, so $lb=8$ and $l=\frac{5b}{4}$. Thus $b^{2}=\frac{32}{5}$, $b=\frac{4\sqrt{10}}{5}\,\mathrm{cm}$ and $l=\sqrt{10}\,\mathrm{cm}$.

Perimeter =$2\left(l+b\right)=\frac{18\sqrt{10}}{5}\,\mathrm{cm}\approx 11.38\,\mathrm{cm}$. Check: the large width is $4\sqrt{10}$, height $\frac{9\sqrt{10}}{5}$; their product is $72\,\mathrm{cm}^{2}$.

Let the trisection points on BC be D,E. The blue triangle ABD and red triangle AEC have equal bases $\mathrm{BD}=\mathrm{EC}$ and the same perpendicular height from A to BC. Each therefore has one-third the area of ABC.

Sample dissection method: transform each triangle to the same rectangle of width BD and height $\frac{h}{2}$. Cut the triangle along the segment joining the midpoints of its two sloping sides. Rotate the small top triangle 180° about one endpoint of this cut and attach it to the remaining trapezium; this makes a parallelogram of base BD and height $\frac{h}{2}$. Cut off the projecting triangular part and translate it to the opposite side to obtain a rectangle. If the parallelogram is very slanted, cut it into narrower horizontal strips first, then perform that operation on each strip. Superimpose the cut lines of the two rectangular arrangements to refine the pieces, then reverse the red-triangle rearrangement. This gives a finite dissection from blue to red; no claim of the fewest pieces is made.

Let the square side be a, and call the quarter-disc area $Q=\frac{\pi a^{2}}{4}$. Each semicircle has radius $\frac{a}{2}$, so the sum of their areas $S_{1}+S_{2}=\frac{\pi a^{2}}{4}=Q$. Their overlap is A; hence their union has area $S_{1}+S_{2}-A$. The region B is the part of the quarter-disc outside that union: $B=Q-\left(S_{1}+S_{2}-A\right)=A$.

A and B are the shaded lunes (crescents), and C is the triangle in Fig. 6.52. They are not three full semicircles.

Let the right-triangle legs be a,b and hypotenuse c. Full semicircle areas on those diameters are Sₐ=$\frac{\pi a^{2}}{8}$, Sᵦ=$\frac{\pi b^{2}}{8}$ and S꜀=$\frac{\pi c^{2}}{8}$. By Pythagoras, Sₐ+Sᵦ=S꜀. Let the two circular segments between the legs and the hypotenuse semicircle be X,Y. Then Sₐ=A+X, Sᵦ=B+Y, while S꜀=C+X+Y. Subtracting X+Y gives $A+B=C$.

Let the rectangle have width w and height h. Take its bottom-left corner as (0,0). Let the interior point be (x,y), with vertically aligned top point (x,h) and horizontally aligned right point (w,y), as drawn. The blue triangle A has area $\frac{x\left(h-y\right)}{2}$, green B has area $\frac{y\left(w-x\right)}{2}$, and pink C has area $\frac{\left(w-x\right)\left(h-y\right)}{2}$.

Therefore $A+C=\frac{w\left(h-y\right)}{2}$ and $B+C=\frac{h\left(w-x\right)}{2}$. Since $C>0$, $\frac{2\left(A+C\right)\left(B+C\right)}{C}=wh$, the rectangle’s area. The horizontal/vertical alignments shown in the figure are essential.

Let $\mathrm{OA}=\mathrm{OB}=r$. Triangle AOB is right-angled with $\mathrm{AB}=r\sqrt{2}$, so its area $T=\frac{r^{2}}{2}$. The quarter-disc AOB has area $Q=\frac{\pi r^{2}}{4}$. The semicircle with diameter AB has radius $\frac{r}{\sqrt{2}}$ and area $$\begin{aligned}S&=\left(\frac{1}{2}\right)\pi \left(\frac{r}{\sqrt{2}}\right)^{2}\\&=\frac{\pi r^{2}}{4}\\&=Q\text{.}\end{aligned}$$

The unshaded circular segment between chord AB and arc AFB has area Q−T. Removing it from the semicircle leaves the shaded lune: $S-\left(Q-T\right)=T$. Thus the lune and the shaded triangle have equal areas.