Important for :

1

2

3

B

In ΔBCF,By the pythagoras theorem,

BF

(BF)

Now, in ΔACF,

AC

AC = 45 cm

Similarly, BD = 45 cm

Hence, option B is correct.

4

B

Area of rectangle lies between 40 cmNow, one side = 5 cm

Since, area can't be less than 40 cm

∴ Other side can't be less than = |
40 | = 8 cm |

5 |

Since, area can't be greater than 45 cm

∴ Other side can't be greater than = |
45 | = 9 cm |

5 |

∴ Minimum value of diagonal = |
8^{2} + 5^{2} |
= | 89 | = 9.43 cm |

∴ Maximum value of diagonal = |
9^{2} + 5^{2} |
= | 106 | = 10.3 cm |

So, diagonal lies between 9 cm and 11 cm.

Hence, option B is correct.

5

I. Area of triangle APS < Area of triangle DSR, if BD < AC.

II. Area of triangle ABC = 4 (Area of triangle BPQ).