So No | Base | Height | Area of parallelogram |
---|---|---|---|
a. | 20 cm | - | 246 \(cm^2\) |
b. | - | 15 cm | 154.5 \(cm^2\) |
c. | - | 8.4 cm | 48.72 \(cm^2\) |
d. | 15.6 cm | - | 16.38 \(cm^2\) |
\(Area\; of\; parallelogram = Base × Height\)
(a) \(b = 20 \;cm, \;h\) = \(?\)
\(Area = 246 \;cm^2\)
\(20 × h = 246\)
\(h\) = \(\frac{246}{20}=12.3 \;cm\)
Therefore, the height of such \(parallelogram\) is \(12.3\) \(cm\).
(b) \(b = ? \;h\)= \(15 \;cm\)
\(Area\) = \(154.5\; cm^2\; b × 15\)
\(b = \frac{154.5}{15}=10.3 \;cm\)
Therefore, the base of such \(parallelogram\) is \(10.3\) \(cm\).
(c) \(b = ? \;h\) = \(8.4\; cm\; Area\) = \(48.72\; cm^2\)
\(b\times h\) = 48.72
= \(b\) = \(\frac{48.72}{8.4}=5.8 \;cm\)
Therefore, the base of such \(parallelogram\) is \(5.8\) \(cm\).
(d) \(b = 15.6\; cm\; h = ?\)
\(Area = 16.38 \;cm^2\)
\(15.6 × h = 16.38\)
\(h= \frac{16.38}{15.6}=1.05\;cm\)
Therefore, the height of such \(parallelogram\) is \(1.05\) \(cm\).
Match the items given in Column I with one or more items of Column II.
Column I | Column II |
(a) A plane mirror | (i) Used as a magnifying glass. |
(b) A convex mirror | (ii) Can form image of objects spread over a large area. |
(c) A convex lens | (iii) Used by dentists to see enlarged image of teeth. |
(d) A concave mirror | (iv) The image is always inverted and magnified. |
(e) A concave lens | (v) The image is erect and of the same size as the object. |
- | (vi) The image is erect and smaller in size than the object. |