Base | Height | Area of triangle |
---|---|---|
15 cm | - | 87 \(cm^2\) |
- | 31.4 mm | 1256 \(mm^2\) |
22 cm | - | 170.5 \(cm^2\) |
\(\text{Area of a triangle}=\frac{1}{2}\times\text{Base}\times\text{Height}\)
(a) \(b = 15 \;cm\)
\(h=?\)
\(Area = \frac{1}{2}\times b\times h=87\; cm^2\)
\(\frac{1}{2}\times15\times h=87\;cm^2\)
\(h= \frac{87\times2}{15}=11.6\;cm^2\)
Therefore, the height of such triangle is \(11.6\) \(cm\).
(b) \(b = ?\)
\(h=31.4\; mm\)
\(Area=\frac{1}{2}\times b\times h=1256 \; mm^2\)
\(b = \frac{1256\times2}{31.4}=80\; mm\)
Therefore, the base of such triangle is \(80 \) \(mm\).
(c) \(b = 22 \;cm ,\;h = ?\)
\(Area=\frac{1}{2}\times b\times h=170.5\; cm^2\)
\(\frac{1}{2}\times 22\times h=170.5\;cm^2\)
\(h=\frac{170.5\times2}{22}=15.5\;cm\)
Therefore, the height of such triangle is \(15.5\) \(cm\).
In triangle \( PQR \), the lengths of \( PT \) and \( TR \) are in the ratio \( 3:2 \). ST is parallel to QR. Two semicircles are drawn with \( PS \) and \( PQ \) as diameters, as shown in the figure. Which one of the following statements is true about the shaded area \( PQS \)? (Note: The figure shown is representative.)
Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30